RD Sharma XI 2018 Solutions for Class 12 Science Math Chapter 29 Limits are provided here with simple step-by-step explanations. These solutions for Limits are extremely popular among class 12 Science students for Math Limits Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XI 2018 Book of class 12 Science Math Chapter 29 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XI 2018 Solutions. All RD Sharma XI 2018 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 29.11:

Question 1:

Show that limx0xx does not exist.

Answer:

limx0 xx

Left hand limit:
limx0- xx Let x=0-h, where h0.limh0 0-h0-h=limh0 -hh=-1

Right hand limit:
limx0+  xxLet x=0+h, where h0.limh0 0+h0+h=limh0 hh=1

Left hand limit ≠ Right hand limit
Thus,lim x0 xx does not exist.

Page No 29.11:

Question 2:

Find k so that limx2fx may exist, where fx=2x+3,x2x+k,x>2.

Answer:

fx=2x+3,x2x+k,x>2Left hand limit:limx2- fx=limx2- 2x+3Let x=2-h, where h0.limh0 22-h+3=22-0+3=7Right hand limit:limx2+ fx=limx2+ x+kLet x=2+h, where h0.limh0 2+h+k=2+k

Now, limx2 fx exists if the left hand limit is equal to the right hand limit.

⇒7 = 2 + k
k = 5

Page No 29.11:

Question 3:

Show that limx0 1x does not exist.

Answer:

limx0 1xLeft hand limit:limx0- 1xLet x=0-h, where h0limh0 10-h=-Right hand limit:limx0+ 1xLet x=0+h, where h0limh0 10+h=limx0- 1xlimx0+ 1xThus, limx0 1x does not exist.

Page No 29.11:

Question 4:

Let f(x) be a function defined by fx=3xx+2x,x0      0,x=0.
Show that limx0  fx does not exist.

Answer:

fx=3xx+2x,x0            0,x=0Left hand limit:limx0- 3xx+2xLet x=0-h, where h0.limh0 3-h-h+2-h=limh0 -3hh-2h=limh0 -3h-h=3Right hand limit:limx0+ 3xx+2xLet x=0+h, where h0.limh0 3hh+2h=limh0 3hh+2h=1 limx0- 3xx+2xlimx0+ 3xx+2xThus, limx0 fx does not exist.

Page No 29.11:

Question 5:

Let fx=x+1,if x0x-1,if x<0.Prove that limx0  fx does not exist.

Answer:

fx=x+1,x0x-1,x<0RHL:limx0+ fx=limx0 x+1Let x=0+h, where h0.limh0 0+h+1=1LHL:limx0- fx=limx0- x-1Let x=0-h, where h0.limh0 0-h-1=-1 LHLRHLThus,limx0 fx does not exist.

Page No 29.11:

Question 6:

Let fx=x+5,if x>0x-4,if x<0. Prove that limx0 fx does not exist.

Answer:

We have,

fx=x+5,if x>0x-4,if x<0

LHL of f(x) at x = 0

= limx0- fx=limh0f0-h=limh00-h-4=-4

RHL of f(x) at x = 0

= limx0+ fx=limh0f0+h=limh00+h+5=5

Clearly, limx0-fxlimx0+fx

Hence, limx0 fx does not exist.

Page No 29.11:

Question 7:

Find limx3 fx, where fx=4,if x>3x+1,if x<3

Answer:

We have,

fx=4,if x>3x+1,if x<3

LHL of f(x) at x = 3

= limx3- fx=limh0f3-h=limh03-h+1=4

RHL of f(x) at x = 3

= limx3+ fx=limh0f3+h=limh04=4

Clearly, limx3-fx=limx3+fx=4

limx3 fx=4

Page No 29.11:

Question 8:

If fx=2x+3,x03 x+1,x>0.Find limx0 fx and limx1 fx.

Answer:

fx=2x+3,x03x+1,x>0LHL:limx0- fx=limx0- 2x+3Let x=0-h, where h0.limh0 20-h+3=limh0 -2h+3=3RHL:limx0+ fxlimx0+ 3x+1Let x=0+h, where h0.limh0 30+h+1=limh0 3h+3=3LHL=RHLlimx0 fx=3

(ii)
limx1 fx=?LHL:limx1- fx=limx1- 3x+1Let x=1-h, where h0.limh0 31-h+1=6RHL:limx1+ fx=limx1+ 3x+1Let x=1+h, where h0.limh0 31+h+1=6LHL=RHLlimx1 fx=6

Page No 29.11:

Question 9:

Find limx1 fx, if fx=x2-1,x1-x2-1,x>1

Answer:

limx1 fx=?fx=x2-1,x1-x2-1,x>1LHL:limx1- fx=limx1- x2-1Let x=1-h, where h0.limh0 1-h2-1=0RHL:limx1+ fxlimx1+ -x2-1Let x=1+h, where h0.limh0 -1+h2-1=-2 LHLRHLThus, limx1 fx does not exist.

Page No 29.11:

Question 10:

Evaluate limx0 fx, where fx=xx,x00,x=0

Answer:

fx=xx,x00,x=0LHL:limx0- xxLet x=0-h, where h0.limh0 0-h-h=limh0 h-h=-1RHL:limx0+ fx=limx0+ xxLet x=0+h, where h0.limh0 0+h0+h=1LHLRHLThus, limx0 fx does not exist.

Page No 29.11:

Question 11:

Let a1, a2, ..., an be fixed real numbers such that
f(x) = (xa1) (xa2) ... (xan)
What islimxa1 fx? For a ≠ a1, a2, ..., an. Compute limxa fx.

Answer:

fx=x-a1 x-a2 ... x-anlimxa1 fx=limxa1 x-a1 x-a2 ... x-an=a1-a1 a1-a2 ... a1-an=0limxa fx=limxa x-a1 x-a2 ... x-an=a-a1 a-a2 ... a-an

Page No 29.11:

Question 12:

Find lim  x1+ 1x-1.

Answer:

limx1+ 1x-1Let x=1+h, where h0.limh0 11+h-1=

Page No 29.11:

Question 13:

Evaluate the following one sided limits:

(i)
limx2+ x-3x2-4

(ii)
limx2- x-3x2-4

(iii)
limx0+ 13x

(iv)
limx-8+ 2xx+8

(v)
limx0+ 2x1/5

(vi)
limxπ2 tan x

(vii)
limx-π/2+sec x

(viii)
limx0- x2-3x+2x3-2x2

(ix)
lim  x-2+ x2-12x+4

(x)
lim x0- 2-cot x

(xi)
limx0- 1+cosec x

Answer:

(i) 
limx2+ x-3x2-4Let x=2+h, where h0.limh0 2+h-32+h2-22=limh0 -1+h2+h-2 2+h+2=-

(ii)
limx2- x-3x2-4Let x=2-h, where h0.limh0 2-h-32-h2-22=limh0 -h-12-h-2 2-h+2=limh0 -h-1-h 4-h=limh01+hh4-h= 

(iii)
limx0+ 13xLet x=0+h, where h0.limh0 13h=

(iv)
limx-8+ 2xx+8Let x=-8+h, where h0.limh0 2-8+h-8+h+8=limh0 -16+2hh=-

(v)
limx0+ 2x15Let x=0+h, where h0.limh0 2h15=

(vi)
 limxπ2- tan xLet x=π2-h, where h0.limh0 tanπ2-h=limh0 cot h=

(vii)
limx-π2+ sec xLet x=-π2+h, where h0.limh0 sec-π2+h=limh0 sec π2-h                            sec-θ=sec θ=limh0 cosec h=

(viii)
 limx0- x2-3x+2x3-2x2=limx0- x2-2x-x+2x2x-1=limx0- xx-2-1x-2x2 x-2=limx0- x-1 x-2x2x-2Let x=0-h, where h0.limh0 -h-1-h2=-

(ix) 
limx-2+ x2-12x+4Let x=-2+h, where h0.limh0 -2+h2-12-2+h+4=4-1-4+4=30=

(x) 
limx0- 2-cot xLet x=0-h, where h0.limh0 2-cot -h=limh0 2+cot h=2+=

(xi)
limx0- 1+cosec xLet x=0-h, where h0.limh0 1+cosec -h=limh0 1-cosec h                    cosec -θ=-cosec θ=1-=-



Page No 29.12:

Question 14:

Show that limx0 e-1/x does not exist.

Answer:

limx0 e-1xLeft hand limit:limx0- e-1xLet x=0-h, where h0.limh0 e-10-h=limh0 e1h=limh0 e1h                 When h0, then 1h.=e=Right hand limit:limx0+ e-1xLet x=0+h, where h0.limh0 e-10+h=limh0 1e1h=1=0 limx0- e-1xlimx0+ e1xThus, limitx0 e-1x does not exist.

Page No 29.12:

Question 15:

Find:

i limx2 x(ii) limx52x(iii) limx1 x

Answer:

(i)

limx2 xLHL:limx2- xLet x=2-h, where h0.limh0 2-h=1RHL:limx2+ xLet x=2+h, where h0.limh0 2+h=2 LHL RHLThus,  lim  x2 x does not exist.


(ii) 
limx52 xLHL limx52- xLet x=52-h, where h0.limh0 52-h=2RHL:limx52+ xLet x=52+h, where h0.limh0 52+h          limx52 x=2=2limx52 x=2

(iii) 
limx1 xLHL:limx1- xLet x=1-h, where h0.limh0 1-h=0RHL:limx1+ xLet x=1+h, where h0.limh0 1+h=1 LHL RHLThus,lim x1 x does not exist.

Page No 29.12:

Question 16:

Prove that lim  xa+ x=a for all aR. Also, prove that lim  x1- x=0.

Answer:

limxa+ xLet x=a+h, where h0.limh0 a+h=alimx1- xLet x=1-h, where h0.limh0 1-h=0

Page No 29.12:

Question 17:

Show that limx2- xxlimx2+xx.

Answer:

limx2- xxLet x=2-h, where h0.limh0 2-h2-h=21limx2+ xxLet x=2+h, where h0.limh0 2+h2+h=22=1 limx2- xxlimx2+xx

Page No 29.12:

Question 18:

Find limx3+ xx. Is it equal to limx3- xx.

Answer:

limx3+ xxLet x=3+h, where h0.limh0 3+h3+h=33=1Also, limx3- xxLet x=3-h, where h0.limh0 3-h3-h=32limx3- xxlimx3+ xx

Page No 29.12:

Question 19:

Find limx5/2 x.

Answer:

limx52 xLHL:limx52- xLet x=52-h, where h0.limh0 52-h=2RHL:limx52+ xLet x=52+h, where h0.limh0 52+h=2

We know:
limx52x=limx52-x=limx52+x

limx52x=2

Page No 29.12:

Question 20:

Evaluate limx2 fx (if it exists), where fx=x-x,x<24,x=23x-5,x>2.

Answer:

fx=x-x,x<24,x=23x-5,x>2LHL:limx2- fx=limx2- x-xLet x=2-h, where h0.limh0 2-h-2-h=2-1=1RHL:limx2+ fx=limx2+ 3x-5Let x=2+h, where h0.limh0 32+h-5=6-5=1Here, LHL=RHL=1limx2fx=1

Page No 29.12:

Question 21:

Show that limx0 sin 1x does not exist.

Answer:

limx0 sin 1xL.H.Llimx0- fxlimx0- sin 1xLet x=0-h where h0=limh0 sin -1h=-limh0 sin 1h

limx0- fx = – (An oscillating number that oscillates between –1 and 1)

Thus,limx0-fx does not exist.Similarly, limx0+fx does not exist.

Page No 29.12:

Question 22:

Let fx=kcosxπ-2x,where xπ23,where x=π2 and if limxπ2 fx=fπ2, find the value of k.

Answer:

We have,

fx=kcosxπ-2x,where xπ23,where x=π2

It is given that,

limxπ2fx=fπ2limxπ2kcosxπ-2x=3k2×limx-π20sinπ2-xπ2-x=3k2×limh0sin-h-h=3                                   Put x-π2=h
k2×limh0sinhh=3                                   sin-θ=-sinθk2=3                                                        limx0sinxx=1k=6

Hence, the value of k is 6.



Page No 29.18:

Question 1:

limx1 x2+1x+1

Answer:

limx1x2+1x+1=12+11+1=1

Page No 29.18:

Question 2:

limx0 2x2+3x+4x2+3x+2

Answer:

limx02x2+3x+4x2+3x+2=2×0+3×0+40+3×0+2=42=2

Page No 29.18:

Question 3:

limx3 2x+3x+3

Answer:

limx32x+3x+3=2×3+33+3=36=12

Page No 29.18:

Question 4:

limx1 x+8x

Answer:

limx1x+8x=1+81=3

Page No 29.18:

Question 5:

limxa x+ax+a

Answer:

limxax+ax+a=a+aa+a=2a2a=1a

Page No 29.18:

Question 6:

limx1 1+x-121+x2

Answer:

limx11+x-121+x2=1+1-121+12=12

Page No 29.18:

Question 7:

limx0 x2/3-9x-27

Answer:

limx0x2/3-9x-27=0-90-27=13

Page No 29.18:

Question 8:

limx0 9

Answer:

limx09=9

f(x) = 9 is a constant function.
Its value does not depend on x.

Page No 29.18:

Question 9:

limx2 3-x

Answer:

limx23-x=3-2=1

Page No 29.18:

Question 10:

limx-1 4x2+2

Answer:

limx-14x2+2=4-12+2=4+2=6

Page No 29.18:

Question 11:

limx-1 x3-3x+1x-1

Answer:

limx-1x3-3x+1x-1=-13-3-1+1-1-1=-1+3+1-2=-32

Page No 29.18:

Question 12:

limx0 3x+1x+3

Answer:

limx03x+1x+3=3×0+10+3=13

Page No 29.18:

Question 13:

limx3 x2-9x+2

Answer:

limx3x2-9x+2=32-93+2=9-95=0

Page No 29.18:

Question 14:

limx0 ax+bcx+d, d 0

Answer:

limx0ax+bcx+d=a×0+bc×0+d=bd



Page No 29.23:

Question 1:

limx-5 2x2+9x-5x+5

Answer:

limx-52x2+9x-5x+5It is of the form 00.limx-52x2+10x-x-5x+5=limx-52xx+5-1x+5x+5=limx-52x-1x+5x+5=limx-52x-1=2-5-1=-11

Page No 29.23:

Question 2:

limx3 x2-4x+3x2-2x-3

Answer:

limx3x2-4x+3x2-2x-3It is of the form 00.limx3x2-x-3x+3x2-3x+x-3=limx3x-3x-1x+1x-3=limx3x-1x+1=3-13+1=12

Page No 29.23:

Question 3:

limx3 x4-81x2-9

Answer:

limx3x4-81x2-9It is of the form 00.limx3x22-92x2-9=limx3x2-9x2+9x2-9=limx3x2+9=32+9=18

Page No 29.23:

Question 4:

limx2 x3-8x2-4

Answer:

limx2x3-8x2-4It is of the form 00.limx2x-2x2+2x+4x-2x+2                  A3-B3=A-BA2+AB+B2A2-B2=A-BA+B=22+2×2+42+2=124=3

Page No 29.23:

Question 5:

limx-1/2 8x3+12x+1

Answer:

limx-1/28x3+12x+1It is of the form 00.limx-1/22x3+12x+1=limx-1/22x+12x2-2x×1+122x+1           A3+B3=A+BA2-AB+B2=limx-1/22x2-2x+1=2×-122-2×-12+1=1+1+1=3

Page No 29.23:

Question 6:

limx4 x2-7x+12x2-3x-4

Answer:

limx4x2-7x+12x2-3x-4It is of the form 00.limx4x2-3x-4x+12x2-4x+x-4=limx4xx-3-4x-3xx-4+1x-4=limx4x-4x-3x-4x+1=4-34+1=15

Page No 29.23:

Question 7:

limx2 x4-16x-2

Answer:

limx2x4-16x-2It is of the form 00.limx2x22-42x-2=limx2x2-4x2+4x-2=limx2x-2x+2x2+4x-2=2+222+4=32

Page No 29.23:

Question 8:

limx5 x2-9x+20x2-6x+5

Answer:

limx5x2-9x+20x2-6x+5It is of the form 00.limx5x2-4x-5x+20x2-x-5x+5=limx5xx-4-5x-4xx-1-5x-1=limx5x-5x-4x-1x-5=5-45-1=14

Page No 29.23:

Question 9:

limx-1 x3+1x+1

Answer:

limx-1x3+1x+1It is of the form 00.limx-1x3+13x+1=limx-1x+1x2-x+1x+1=-12--1+1=1+1+1=3

Page No 29.23:

Question 10:

limx5 x3-125x2-7x+10

Answer:

limx5x3-125x2-7x+10It is of the form 00.limx5x3-53x2-2x-5x+10=limx5x-5x2+5x+52xx-2-5x-2      A3-B3=A-BA2+AB+B2=limx5x-5x2+5x+25x-2x-5=52+5×5+255-2=753=25

Page No 29.23:

Question 11:

limx2 x2-2x2+2x-4

Answer:

limx2x2-2x2+2x-4It is of the form 00.limx2x2-22x2+22x-2x-4=limx2x-2x+2xx+22-2x+22=limx2x-2x+2x-2x+22=2+22+22=2232=23

Page No 29.23:

Question 12:

limx3 x2-3x2+3 3x-12

Answer:

limx3x2-3x2+33x-12It is of the form 00.limx3x2-32x2+43x-3x-12=limx3x-3x+3xx+43-3x+43=limx3x-3x+3x-3x+43=3+33+43=25

Page No 29.23:

Question 13:

limx3 x4-9x2+43x-15

Answer:

limx3x4-9x2+43x-15It is of the form 00.limx3x22-32x2+53x-3x-15=limx3x2-3x2+3xx+53-3x+53=limx3x2-32x2+3x-3x+53=limx3x-3x+3x2+3x-3x+53=3+33+33+53=23×663=2

Page No 29.23:

Question 14:

limx2 xx-2-4x2-2x

Answer:

limx2xx-2-4x2-2x=limx2xx-2-4xx-2=limx2x2-4xx-2=limx2x-2x+2xx-2=limx2x+2x=2+22=2

Page No 29.23:

Question 15:

limx1 1x2+x-2-xx3-1

Answer:

 limx1x3-1-xx2+x-2x2+x-2x3-1=limx1x3-1-x3-x2+2xx2+x-2x3-1=limx1-x2+2x-1x2+x-2x-1x2+x+1=limx1-x2-2x+1x2+x-2x-1x2+x+1=limx1-x-12x2+x-2x-1x2+x+1=limx1-x-1x2+2x-x-2x2+x+1=limx1-x-1xx+2-1x+2x2+x+1=limx1-x-1x-1x+2x2+x+1=-11+21+1+1=-19

Page No 29.23:

Question 16:

limx3 x4-9x2+43x-15

Answer:

limx31x-3-2x2-4x+3=limx31x-3-2x2-3x-x+3=limx31x-3-2xx-3-1x-3=limx31x-3-2x-1x-3=limx3x-1-2x-3x-1=limx31x-1=13-1=12

Page No 29.23:

Question 17:

limx2 1x-2-2x2-2x

Answer:

limx21x-2-2x2-2x=limx21x-2-2xx-2=limx2x-2xx-2=limx21x=12

Page No 29.23:

Question 18:

limx1/4 4x-12x-1

Answer:

limx1/44x-12x-1It is of the form 00.limx1/42x2-122x-1=limx1/42x-12x+12x-1=214+1=2×12+1=2

Page No 29.23:

Question 19:

limx4 x2-16x-2

Answer:

limx4x2-16x-2It is of the form 00.limx4x2-42x-2=limx4x-4x+4x-2=limx4x2-22x+4x-2=limx4x-2x+2x+4x-2=2+24+4=32

Page No 29.23:

Question 20:

limx0 a+x2-a2x

Answer:

limx0a+x2-a2xIt is of the form 00.limx0a2+x2+2ax-a2x=limx0xx+2ax=limx0x+2a=0+2a=2a

Page No 29.23:

Question 21:

limx2 1x-2-4x3-2x2

Answer:

limx21x-2-4x3-2x2=limx21x-2-4x2x-2=limx2x2-4x2x-2=limx2x-2x+2x2x-2=limx2x+2x2=2+222=1

Page No 29.23:

Question 22:

limx3 1x-3-3x2-3x

Answer:

limx31x-3-3x2-3x=limx31x-3-3xx-3=limx3x-3xx-3=limx31x=13

Page No 29.23:

Question 23:

limx1 1x-1-2x2-1

Answer:

limx11x-1-2x2-1=limx11x-1-2x-1x+1=limx1x+1-2x-1x+1=limx1x-1x-1x+1=limx11x+1=11+1=12

Page No 29.23:

Question 24:

limx3 x2-9 1x+3+1x-3

Answer:

limx3x2-91x+3+1x-3=limx3x2-9x-3+x+3x+3x-3=limx3x2-92xx2-9=limx32x=2×3=6

Page No 29.23:

Question 25:

limx1 x4-3x3+2x3-5x2+3x+1

Answer:

p(x) = x4 - 3x3 + 2
p(1) = 1 - 3 + 2
       = 0
Now, x-1 is a factor of p(x).

          x3-2x2-2x-2x-1x4-3x3+2                         x4-x3       -    +                 -2x3+2           -2x3+2x2           +    -                           - 2x2+2                    - 2x2+2x                     +      -                               -2x+2                        -2x+2                        +     -                     ___________                            0

q(x) = x3 - 5x2 + 3x + 1
q(1) = 1 - 5 + 3 + 1
       = 0
Now, x-1 is a factor of q(x).

          x2-4x-1x-1x3-5x2+3x+1          x3-x2       -    +                     4x2+3x+1               4x2+4x           -      -                            -x+1                     -x+1                     +   -                               0   


limx1x4-3x3+2x3-5x2+3x+1=limx1x-1x3-2x2-2x-2x-1x2-4x-1=(1)3-212-21-212-4×1-1=1-2-2-21-4-1=-5-4=54

Page No 29.23:

Question 26:

limx2 x3+3x2-9x-2x3-x-6

Answer:

limx2x3+3x2-9x-2x3-x-6

It is of the form 00.

Let p(x) = x3 + 3x2 - 9x - 2
p(2) = 8 + 12 - 18 - 2
       = 0
Now, x-2 is a factor of p(x).

          x2+5x+1x-2x3+3x2-9x-2          x3-2x2       -    +                     5x2-9x-2               5x2-10x           -      +                                     x-2                         x-2                       - +                                 0   

Let q(x) = x3 – x – 6
q(2) = 8 – 2 – 6
       = 0
Now, x-2 is a factor of q(x).

          x2+2x+3x-2x3-x-6              x3-2x2       -    +                     2x2-x-6               2x2-4x           -      +                               3x-6                        3x-6                     -    +                               ×××    


limx2x3+3x2-9x-2x3-x-6=limx2x-2x2+5x+1x-2x2+2x+3=limx2x2+5x+1x2+2x+3=(2)2+5×2+122+2×2+3=4+10+14+4+3=1511

Page No 29.23:

Question 27:

limx1 1-x-1/31-x-2/3

Answer:

limx11-x-1/31-x-2/3It is of the form 00.limx11-x-1/312-x-1/32=limx11-x-1/31-x-131+x-1/3=limx111+x-1/3=11+1=12

Page No 29.23:

Question 28:

limx3 x2-x-6x3-3x2+x-3

Answer:

limx3x2-x-6x3-3x2+x-3It is of the form 00.limx3x2-3x+2x-6x2x-3+1x-3=limx3xx-3+2x-3x2+1x-3=limx3x+2x-3x2+1x-3=3+232+1=510=12

Page No 29.23:

Question 29:

limx-2 x3+x2+4x+12x3-3x+2

Answer:

Let p(x) = x3 + x2 + 4x + 12
p(–2) = 0
Thus, x = –2 is the root of p(x).
Now, x+2 is a factor of p(x).

          x2 - x + 6x+2x3 + x2 + 4x + 12          x3 + 2x2       -    -             -x2 + 4x + 12       -x2 - 2x       +     +                         6x + 12                   6x + 12                -     -                                0       

p(x) = x3 + x2 + 4x + 12
       = (x + 2)(x2x + 6)

Let q(x) = x3 – 3x + 2
q-2 = -8 + 6 + 2
         = 0
Thus, x = -2 is the root of q(x).
Now, x+2 is a factor of q(x).

          x2-2x+1x+2x3-3x+2          x3+2x2                        -2x2-3x+2       -2x2-4x       +      +                         x+2                   x+2                                           ×××  

q(x) = (x + 2)(x2 – 2x + 1)

limx-2x3+x2+4x+12x3-3x+2=limx-2x+2x2-x+6x+2x2-2x+1=(2)2--2+6-22-2-2+1=4+2+64+4+1=129=43

Page No 29.23:

Question 30:

limx1 x3+3x2-6x+2x3+3x2-3x-1

Answer:

Let p(x) = x3 + 3x2 - 6x + 2
p(1) = 1 + 3 - 6 + 2
       = 0
Now, x+2 is a factor of p(x).

          x2+4x-2x-1x3+3x2-6x+2          x3-x2       -    +                 -4x2-6x+2           -4x2-4x           +    +                           - 2x+2                    - 2x+2                     +     -                                   0         

p(x) = (x - 1)(x2 + 4x - 2)

q(x) = x3 + 3x2 - 3x + 2
q(1) = 1 + 3 - 3 - 1
       = 0
Now, x+2 is a factor of p(x).

          x2+4x+1 x-1x3+3x2-3x-1          x3-x2       -    +                     4x2-3x-1               4x2-4x           -      +                                 x-1                          x-1                     -    +                                   0        


limx1x3+3x2-6x+2x3+3x2-3x-1=limx1x-1x2+4x-2x-1x2+4x+1=(1)2+4×1-212+4×1+1=1+4-21+4+1=36=12



Page No 29.24:

Question 31:

limx2 1x-2-22x-3x3-3x2+2x

Answer:

limx21x-2-22x-3x3-3x2+2x=limx21x-2-22x-3xx2-3x+2=limx21x-2--22x-3xx2-2x-x+2=limx21x-2-22x-3xxx-2-1x-2=limx21x-2-22x-3xx-1x-2=limx2xx-1-22x-3xx-1x+2=limx2x2-x-4x+6xx-1x+2=limx2x2-5x+6xx-1x+2=limx2x2-2x-3x+6xx-1x-2=limx2xx-2-3x-2xx-1x-2=limx2x-3x-2xx-1x-2=2-322-1=-12

Page No 29.24:

Question 32:

limx1 x2-1+x-1x2-1, x>1

Answer:

limx1x2-1+x-1x2-1It is of the form 00.limx1x2-1x2-1+x-1x2-1=limx11+x-1x-1x+1=1+11+1=1+12=2+12

Page No 29.24:

Question 33:

limx1 x-2x2-x-1x3-3x2+2x

Answer:

limx1x-2x2-x-1x3-3x2+2x=limx1x-2xx-1-1xx2-3x+2=limx1x-2xx-1-1xx2-2x-x+2=limx1x-2xx-1-1xxx-2-1x-2
=limx1x-2xx-1-1xx-1x-2=limx1x-22-1xx-1x-2

=limx1x-1x-3xx-1x-2=1-311-2=-2-1=2

Page No 29.24:

Question 34:

Evaluate the following limits:

limx1x7-2x5+1x3-3x2+2

Answer:

When x = 1, the expression x7-2x5+1x3-3x2+2 assumes the form 00. So, (x − 1) is a factor of numerator and denominator.

Using long division method, we get

x7-2x5+1=x-1x6+x5-x4-x3-x2-x-1

and x3-3x2+2=x-1x2-2x-2

limx1x7-2x5+1x3-3x2+2=limx1x-1x6+x5-x4-x3-x2-x-1x-1x2-2x-2=limx1x6+x5-x4-x3-x2-x-1x2-2x-2=1+1-1-1-1-1-11-2-2=-3-3=1



Page No 29.28:

Question 1:

limx0 1+x+x2-1x

Answer:

limx01+x+x2-1x

When x = 0, the expression 1+x+x2-1x takes the form 00.
Rationalising the numerator:

limx01+x+x2-11+x+x2+1x1+x+x2+1=limx01+x+x2-1x1+x+x2+1=limx0x1+xx1+x+x2+1=1+01+0+0+1=12

Page No 29.28:

Question 2:

limx0 2xa+x-a-x

Answer:

limx02xa+x-a-x

When x = 0, then the expression 2xa+x-a-x becomes 00.

Rationalising the denominator:

limx02xa+x-a-x×a+x+a-xa+x+a-x

limx02xa+x+a-xa+x-a-x

limx02xa+x+a-x2x

a+a
2a

Page No 29.28:

Question 3:

limx0 a2+x2-ax2

Answer:

limx0a2+x2-ax2

On putting x = 0 in the expression a2+x2-a, it becomes 00.
Rationalising the numerator:

limx0a2+x2-aa2+x2+ax2a2+x2+a

limx0a2+x2-a2x2a2+x2+a

1a2+a

12a

Page No 29.28:

Question 4:

limx0 1+x-1-x2x

Answer:

limx01+x-1-x2x

It is of the form 00.

Rationalising the numerator:

 limx01+x-1-x1+x+1-x2x1+x+1-x

limx01+ x-1-x2x1+x+1-x

limx02x2x1+x+1-x

11+0+1-0
12

Page No 29.28:

Question 5:

limx2 3-x-12-x

Answer:

limx23-x-12-x

It is of the form 00.
Rationalising the numerator:

 limx23-x-13-x+12-x3-x+1

=limx23-x-12-x3-x+1

=limx22-x2-x3-x+1

=13-2+1=11+1=12

Page No 29.28:

Question 6:

limx3 x-3x-2-4-x

Answer:

limx3x-3x-2-4-x

It is of form 00.

Rationalising the denominator:

limx3x-3x-2+4-xx-2-4-xx-2+4-x

limx3x-3x-2+4-xx-2-4-x

limx3x-3x-2+4-x2x-6

limx3x-3x-2+4-x2x-3

3-2+4-32

1+12
22=1

Page No 29.28:

Question 7:

limx0 x1+x-1-x

Answer:

limx0x1+x-1-x

It is of the form 00.

Rationalising the denominator:

 limx0x1+x-1-x×1+x+1-x1+x+1-x

limx0x1+x+1-x1+x-1-x

limx0x1+x+1-x2x

1+12

22
= 1

Page No 29.28:

Question 8:

limx1 5x-4-xx-1

Answer:

limx15x-4-xx-1

It is of the form 00.
Rationalising the numerator:

limx15x-4-xx-15x-4+x5x-4+x

limx15x-4-xx-15x-4+x

limx14x-1x-15x-4+x

45-4+1

42

= 2

Page No 29.28:

Question 9:

limx1 x-1x2+3-2

Answer:

limx1x-1x2+3-2


It is of the form 00.
Rationalising the denominator:

limx1x-1x2+3+2x2+3-2x2+3+2

limx1x-1x2+3+2x2+3-4

limx1x-1x2+3+2x2-1

limx1x-1x2+3+2x-1x+1

=1+3+21+1

42

= 2

Page No 29.28:

Question 10:

limx3 x+3-6x2-9

Answer:

limx3x+3-16x2-9

It is of the form 00.
Rationalising the numerator:

limx3x+3-6x+3+6x2-9x+3+6

limx3x+3-6x-3x+3x+3+6

16×26

1126

Page No 29.28:

Question 11:

limx1 5x-4-xx2-1

Answer:

limx15x-4-xx2-1

It is of the form 00.
Rationalising the numerator:

 limx15x-4-x5x-4+x5x-4+xx2-1

limx15x-4-x5x-4+xx-1x+1

limx14x-15x-4+xx-1x+1

45-4+11+1

42×2

= 1

Page No 29.28:

Question 12:

limx0 1+x-1x

Answer:

limx01+x-1x

It is of the form 00.
Rationalising the numerator:

limx01+x-11+x+1x1+x+1

=limx01+x-1x1+x+1

11+0+1

12

Page No 29.28:

Question 13:

limx2 x2+1-5x-2

Answer:

limx2x2+1-5x-2

It is of the form 00.
Rationalising the numerator:

 limx2x2+1-5x2+1+5x-2x2+1+5

limx2x2+1-5x-2x2+1+5

limx2x2-4x-2x2+1+5

limx2x-2x+2x-2x2+1+5

425

25

Page No 29.28:

Question 14:

limx2 x-2x-2

Answer:

limx2x-2x-2

It is of the form 00.

limx2x2-22x-2

limx2x-2x+2x-2

2+2

22

Page No 29.28:

Question 15:

limx7 4-9+x1-8-x

Answer:

limx74-9+x1-8-x

It is of the form 00.
Rationalising the numerator and the denominator:

limx74-9+x 1×4+9+x4+9+x×11-8-x×1+8-x1+8-x

limx716-9+x4+9+x×1+8-x1-8-x

limx7-1-7+x1+8-x4+9+x-7+x

limx7-1+8-x4+9+x

-1+8-74+9+7

-24+4

-14

Page No 29.28:

Question 16:

limx0 a+x-axa2+ax

Answer:

limx0a+x-axa2+ax

It is of the form 00.
Rationalising the numerator:

limx0a+x-a × a+x+axa2+ax × a+x+a

limx0a+x-axa2+ax a+x+a

1a2 a+a

12aa

Page No 29.28:

Question 17:

limx5 x-56x-5-4x+5

Answer:

limx5x-56x-5-4x+5

It is of the form 00.
Rationalising the denominator:

limx5x-5 6x-5+4x+56x-5-4x+5 6x-5+4x+5

limx5x-5 6x-5+4x+56x-5-4x+5

limx5x-5 6x-5+4x+52x-10

limx5x-56x-5+4x+52x-5

6×5-5+4×5+52

5+52

= 5

Page No 29.28:

Question 18:

limx1 5x-4-xx3-1

Answer:

limx15x-4-xx3-1

It is of the form 00.
Rationalising the numerator:

limx15x-4-x 5x-4+xx3-1 5x-4+x

limx15x-4-xx-1x2+x+15x-4+x

limx14x-1x-1x2+x+15x-4+x

431+1

23

Page No 29.28:

Question 19:

limx2 1+4x-5+2xx-2

Answer:

limx21+4x-5+2xx-2

It is of the from 00.
Rationalising the numerator:

 limx21+4x-5+2x 1+4x+5+2xx-2 1+4x+5+2x

limx21+4x-5+2xx-21+4x+5+2x

limx22x-2x-21+4x+5+2x

21+4×2+5+2×2

23+3

13

Page No 29.28:

Question 20:

limx1 3+x-5-xx2-1

Answer:

limx13+x-5-xx2-1

It is of the form 00.
Rationalising the numerator:

 limx13+x-5-x3+x+5-xx-1x+13+x+5-x

limx13+x-5-xx-1x+13+x+5-x

limx12x-2x-1x+1 3+x+5-x

limx12x-1x-1x+13+x+5-x

22 4+4

14



Page No 29.29:

Question 21:

limx0 1+x2-1-x2x

Answer:

limx01+x2-1-x2x

It is of the form 00.
Rationalising the numerator:

limx01+x2-1-x21+x2+1-x21+x2+1-x2x

limx01+x2-1-x2x1+x2+1-x2

limx02x2x1+x2+1-x2

2×01+0+1-0

= 0

Page No 29.29:

Question 22:

limx0 1+x+x2-x+12x2

Answer:

limx01+x+x2-x+12x2

It is of the form 00.
Rationalising the numerator:

 limx01+x+x2-x+11+x+x2+x+11+x+x2+x+12x2

limx01+x+x2-x+11+x+x2+x+12x2

limx0x21+x+x2+x+12x2

11+0+0+0+1×2

12×12

14

Page No 29.29:

Question 23:

limx4 2-x4-x

Answer:

limx42-x4-x

limx42-x2-x2

limx42-x2-x2+x

12+4

12+2

14

Page No 29.29:

Question 24:

limxa x-ax-a

Answer:

limxax-ax-a

limxax2-a2x-a

limxax-a x+ax-a

a+a

2a

Page No 29.29:

Question 25:

limx0 1+3x-1-3xx

Answer:

limx0 1+3x-1-3xx

It is of the form 00.

Rationalising the numerator:

limx01+3x-1-3x 1+3x+1-3xx 1+3x+1-3x=limx0 1+3x-1-3xx1+3x+1-3x=limx0 6xx 1+3x+1-3x=61+1=62=3

Page No 29.29:

Question 26:

limx0 2-x-2+xx

Answer:

limx02-x-2+xx

It is of the form 00.
Rationalising the numerator:

 limx02-x-2+x2-x+2+xx2-x+2+x

limx02-x-2+xx2-x+2+x

limx0-2xx2-x+2+x

-222

-12

Page No 29.29:

Question 27:

limx1 3+x-5-xx2-1

Answer:

limx1 3+x-5-xx2-1

It is of the form 00.

Rationalising the numerator:

limx1 3+x-5-x 3+x + 5-xx-1 x+1 3+x+5-x=limx13+x-5-xx-1 x+1 3+x +5-x=limx12x-1x-1 x+1 3+x +5-x=21+1 3+1+5-1=22×2+2=14

Page No 29.29:

Question 28:

limx1 2x-3 x-13x2+3x-6

Answer:

limx12x-3x-13x2+3x-6

It is of the form 00.

⇒ limx12x-3x-13x2+x-2

limx12x-3x-13x2+2x-x-2

limx12x-3x-13xx+2-1x+2

limx12x-3x-13x-1x+2

limx12x-3x-13x2-12x+2

limx12x-3x-13x+1x-1x+2

-132×3

-118

Page No 29.29:

Question 29:

limx0 1+x2-1+x1+x3-1+x

Answer:

limx0 1-x2-1+x1+x3-1+x

It is of the form 00.
Rationalising the numerator and the denominator:

limx01+x2-1+x1×1+x2+1+x1+x2+1+x×11+x3-1+x×1+x3-1+x1+x3+1+x=limx0 1+x2-1+x1+x3-1+x×1+x3+1+x1+x2+1+x=limx0x2-xx3-x×1+x3+1+x1+x2+1+x=limx0xx-1xx2-1 1+x3+1+x1+x2+1+x=limx0 x-1 1+x3+1+xx-1 x+1 1+x2+1+x=1+0+1+00+1 1+0+1+0=22=1

Page No 29.29:

Question 30:

limx1 x2-xx-1

Answer:

limx1x2-xx-1

It is of the form 00.

limx1xxx-1x-1

limx1xx3/2-1x12-1

limx1xx3-13x-1            A3-B3=A-BA2+AB+B2

limx1xx-1x+x+1x-1

= 1 (1 + 1 + 1)

= 3

Page No 29.29:

Question 31:

limh0 x+h-xh, x0

Answer:

limh0x+h-xh

It is of the form 00.
Rationalising the numerator:
limh0x+h-xh×x+h+xx+h+x

limh0x+h-xhx+h+x

1x+x

12x

Page No 29.29:

Question 32:

limx10 7+2x-5+2x2-10

Answer:

limx107+2x-5+2x2-102

limx107+2x-5+22x-10x+10

limx107+2x-5+2+252x-10x+10

limx107+2x-7+210x-10x+10

Rationalising the numerator:

 limx107+2x-7+210 7+2x+7+210x-10x+10 7+2x+7+210

limx107+2x-7+210x-10x+10 7+2x+7+210

limx102x-10x-10x+10 7+2x+7+210

210+10 27+210


=2210×27+210=12105+22

12105+2×5-25-2

=12105-252-22

5-2610

Page No 29.29:

Question 33:

limx6 5+2x-3+2x2-6

Answer:

limx65+2x-3+2x2-6

limx65+2x-3+22x2-62

limx65+2x-3+2+26x-6x+6

limx65+2x-5+26x-6x+6

Rationalising the numerator:

limx65+2x-5+265+2x+5+26x-6x+65+2x+5+26

limx65+2x-5+26x-6x+65+2x+5+26

limx62x-6x-6x+65+2x+5+26

26+65+26+5+26

1263+22

1263+2

1263+2×3-23-2

3-2263-2

3-226

Page No 29.29:

Question 34:

limx2 3+2x-2+1x2-2

Answer:

limx23+2x-2+1x2-2

limx23+2x-2+12x-2x+2

limx23+2x-2+1+22x-2x+2

limx23+2x-3+22x-2x+2

Rationalising the numerator:

 limx23+2x-3+223+2x+3+22x-2x+23+2x+3+22

limx23+2x-3+22x-2x+23+2x+3+22

limx22x-2x-2x+23+2x+3+22

22+23+22+3+22

22223+22

1223+22

1222+12

1222+1×2-12-1

2-1222-1

2-122



Page No 29.33:

Question 1:

limxa x+25/2-a+25/2x-a

Answer:

limxa x+252-a+252x-a=limxa x+252-a+252x+2-a+2

Let y = x + 2 and b = a + 2.

When x → a, then x + 2 ​→ a + 2.
 y ​→ b

limyby52-b52y-b=52b52-1=52b32=52a+232

Page No 29.33:

Question 2:

limxa x+23/2-a+23/2x-a

Answer:

limxa=x+232-a+232x-a=limxa x+232-a+232x+2-a+2

Let y = x + 2 and b = a + 2.

When x ​→ a and x + 2 ​→ a + 2.
y ​→ b

limyb y32-b32y-b=32 b32-1=32b12=32 a+212

Page No 29.33:

Question 3:

limx0 1+x6-11+x2-1

Answer:

limx0 1+x6-11+x2-1=limx0 1+x6-1x×x1+x2-1=limx0 1+x6-161+x-1×1+x-11+x2-1

Let y = 1 + x

When x ​→ 0, then 1 + x → 1.
y ​→ 1

limy1y6-16y-1×y-1y2-12=6×16-12×12-1=62=3

Page No 29.33:

Question 4:

limxa x2/7-a2/7x-a

Answer:

limxa x27-a27x-a=27 a27-1                            limxa xn-anx-a=nan-1=27a-57

Page No 29.33:

Question 5:

limxa x5/7-a5/7x2/7-a2/7

Answer:

limxa x57-a57x27-a27=limxa x57-a57x-a×x-ax27-a27=57 a57-1×127 a27-1=52a-27a57=52a-27+57=52a37

Page No 29.33:

Question 6:

limx-1/2 8x3+12x+1

Answer:

limx-12 8x3+12x+1=limx-12 2x3--12x--1When x-12, then 2x → –1.
Let y = 2x

limy-1 y3--13y--1=3-13-1=3×-12=3

Page No 29.33:

Question 7:

limx27 x1/3+3 x1/3-3x-27

Answer:

limx27 x13+3 x13-3x-27=limx27 x13+3 x13-3x133-33 x27 x133Let y=x13limy3y+3 y-3y3-33=3+33×33-1=63×9=29

Page No 29.33:

Question 8:

limx4 x3-64x2-16

Answer:

limx4 x3-64x2-16=limx4 x3-64x-4×x-4x2-16=limx4 x3-43x-4×x-4x2-42=343-1×1242-1=3×162×4=6

Page No 29.33:

Question 9:

limx1 x15-1x10-1

Answer:

limx1 x15-1x10-1=limx1 x15-115x-1×x-1x10-110=15115-110110-1=32

Page No 29.33:

Question 10:

limx-1 x3+1x+1

Answer:

limx-1 x3+1x+1=limx-1 x3--1x--1=limx-1 x3--13x--1=3-13-1=3

Page No 29.33:

Question 11:

limxa x2/3-a2/3x3/4-a3/4

Answer:

limxa x23-a23x34-a34=limaa x23-a23x-a×x-ax34-a34=23a23-1×134a34-1=89 a-13a-14=89a-13+14=89a-112

Page No 29.33:

Question 12:

If limx3 xn-3nx-3=108, find the value of n.

Answer:

limx3 xn-3nx-3=108

x(3)n – 1 = 108
x(3)n – 1 = 4 × 33

On comparing LHS and RHS, we observe that x is equal to 4.

21082543273933 1

Page No 29.33:

Question 13:

If limxa x9-a9x-a=9, find all possible values of a.

Answer:

limxa x9-a9x-a=99a8=9a8=1a=±1

Page No 29.33:

Question 14:

If limxa x5-a5x-a=405, find all possible values of a.

Answer:

limxa x5-a5x-a=4055a4=405a4=81a2=9a=±3

Page No 29.33:

Question 15:

If limxa x9-a9x-a=limx5 4+x, find all possible values of a.

Answer:

limxa x9-a9x-a=limx5 4+x9a9+1=9a8=1a=±1

Page No 29.33:

Question 16:

If limxa x3-a3x-a=limx1 x4-1x-1, find all possible values of a.

Answer:

limxa x3-a3x-a=limx1 x4-14x-13a3-1=414-13a2=4a2=43a=±23



Page No 29.38:

Question 1:

limx 3x-1 4x-2x+8 x-1

Answer:

limx3x-1 4x-2x+8 x-1Dividing the numerator and the denominator by x2:limx 3x-1x 4x-2xx+8x x-1x=limx 3-1x 4-2x1+8x 1-1xWhen x , then 1x0.3×41=12

Page No 29.38:

Question 2:

limx 3x3-4x2+6x-12x3+x2-5x+7

Answer:

limx 3x3-4x2+6x-12x3+x2-5x+7Dividing the numerator and the denominator by x3:limx 3-4x+6x2-1x32+1x-5x2+7x3When x, then 1x, 1x2, 1x30.32

Page No 29.38:

Question 3:

limx 5x3-69+4x6

Answer:

limx 5x3-69+4x6Dividing the numerator and the denominator by x:limx5x3-6x39+4x6x3=limx 5-6x39x6+4=54=52

Page No 29.38:

Question 4:

limx x2+cx-x

Answer:

limx x2+cx-x

It is of the form ∞ - ∞.

Rationalising the numerator:

limxx2+cx-x x2+cx+xx2+cx+x=limxx2+cx-x2x2+cx+xDividing the numerator and the denomiator by x:limx cxxx2+cx+xx=limx c1+cx+1When x, then 1x0.c1+1=c2

Page No 29.38:

Question 5:

limxx+1-x

Answer:

limx x+1-x

It is of the form ∞–​​∞.

On rationalising, we get:
limx x+1-x×x+1+xx+1+x=limx x+1-xx+1+x=1=0

Page No 29.38:

Question 6:

limxx2+7x-x

Answer:

limx x2+7x-xRationalising the numerator:limx x2+7x-x x2+7x+xx2+7x+x=limx x2+7x-x2x2+7x+xDividing the numerator and the denominator by x:limx7x2+7xx+1=limx 7x2+7xx+1When x, then 1x0.71+1=72

Page No 29.38:

Question 7:

limxx4x2+1-1

Answer:

limx x4x2+1-1Rationalising the denominator:limxx4x2+1-1 4x2+1+14x2+1+1=limx x4x2+1+14x2+1-1=limx4x2+1+14xDividing the numerator and the denominator by x:limx 4x2+1x+1x4=limx4x2+1x2+1x4=limx 4+1x2+1x4 x 1x, 1x20=44=24=12

Page No 29.38:

Question 8:

limnn21+2+3+...+n

Answer:

limn n21+2+3.....nIt is of the form .limn n2nn+12=limn2nn+1Dividing the numerator and the denominator by n:limn21+1n=2



Page No 29.39:

Question 9:

limx3x-1+4x-25x-1+6x-2

Answer:

limx 3x-1+4x-25x-1+6x-2=limx 3x+4x25x+6x2=limx3x+45x+6Dividing the numerator and the denominator by x:limx 3+4x5+6x=35

Page No 29.39:

Question 10:

limxx2+a2-x2+b2x2+c2-x2+d2

Answer:

limx x2+a2-x2+b2x2+c2-x2+d2Rationalising the numerator and the denominator:limxx2+a2-x2+b2x2+c2-x2+d2×x2+c2+x2+d2x2+c2+x2+d2×x2+a2+x2+b2x2+a2+x2+b2=limxx2+a2-x2+b2 x2+a2+x2+b2 x2+c2+x2+d2x2+c2-x2+d2 x2+c2+x2+d2 x2+a2+x2+b2=limx x2+a2-x2+b2x2+c2-x2+d2×x2+c2+x2+d2x2+a2+x2+b2=limx a2-b2c2-d2 x2+c2+x2+d2x2+a2+x2+b2Dividing the numerator and the denominator by x:limxa2-b2c2-d2 1+c2x2+1+d2x21+1x2+1+b2x2As x, 1x, 1x20=a2-b2c2-d2 1+11+1=a2-b2c2-d2

Page No 29.39:

Question 11:

limnn+2!+n+1!n+2!-n+1!

Answer:

limnn+2!+n+1!n+2!-n+1!=limn n+2 n+1!+n+1!n+2 n+1!-n+1!=limn n+1!n+1!×n+2+1n+2-1=limn n+3n+1

Dividing the numerator and the denominator by n:

limn 1+3n1+1n When n, then 1n0.11=1

Page No 29.39:

Question 12:

limxxx2+1-x2-1

Answer:

limxxx2+1-x2-1Rationalising the numerator:limxx x2+1-x2-1 x2+1+x2-1x2+1+x2-1=limxx x2+1-x2-1x2+1+x2-1=limxx×2x2+1+x2-1Dividing the numerator and the denominator by x:limx 2x2+1x2+x2-1x2=limx2x2+1x2+x2-1x2=limx21+1x2+1-1x2=21+1=2

Page No 29.39:

Question 13:

limxx+1-x x+2

Answer:

limxx+1-xx+2Rationalising the numerator:limxx+2 x+1-x x+1+xx+1+xlimxx+2 x+1-xx+1+xDividing the numerator and the denominator by x:limx x+2xx+1+xx=limx 1+2x1+1x+1=11+1=12

Page No 29.39:

Question 14:

limn12+22+...+n2n3

Answer:

=limn12+22+... +n2n3=limn nn+1 2n+16n3=limn n+1 2n+16n2=limnn+1n 2n+1n×16 =limn 1+1n 2+1n×16=26=13

Page No 29.39:

Question 15:

limn1+2+3......n-1n2

Answer:

limn1+2+3+...n-1n2limnnn-12 n2limn1-1n×12When n , then 1n0.=12

Page No 29.39:

Question 16:

limn13+23+....n3n4

Answer:

limn 13+23+... +n3n4=limnnn+122n4=limnn2n+124n4=limnn2n+124n4=limn n+124n2=limn n+1n2×14=limn 1+1n2×14When n, then 1n0.14

Page No 29.39:

Question 17:

limn13+23+...n3n-14

Answer:

limn 13+23+... +n3n-14=limn nn+122n-14=limn n2n+124n-14

Dividing the numerator and the denominator by n4:

limn n2n+12n44n-14n4=limnn+12n24n-1n4=limn 1+1n241-1n4When n , then 1n0.1+0241-04=14

Page No 29.39:

Question 18:

limxxx+1-x

Answer:

limxxx+1-x=limx xx+1-xx+1+xx+1+x=limx xx+1-xx+1+x

Dividing the numerator and the denominator by x:

limx1x+1+xx=limx11+1x+1As x, 1x0=11+0+1=12

Page No 29.39:

Question 19:

limn13+132+133+...+13n

Answer:

limn13+132+133+...13n=limn 131+13+132+.....13n-1=limn 13 1-13n1-13               a+ar+ar2+......arn-1=a1-rn1-r=limn 131-13n23=limn12 1-13nAs n, 3n, 13n0=121-0=12

Page No 29.39:

Question 20:

limxx4+7x3+46x+ax4+6, where a is a non-zero real number.

Answer:

limxx4+7x3+46x+ax4+6

Dividing the numerator and the denominator by x4:

limx 1+7x+46x2+9x41+6x4As x, 1x, 1x2, 1x3, 1x40=11

Page No 29.39:

Question 21:

fx=ax2+bx2+1, limx0 fx=1 and limxfx=1, then prove that f(−2) = f(2) = 1

Answer:

fx=ax2+bx2+1limx0 fx=1limx0 ax2+bx2+1=1a×0+ba+1=1b=1Also, limx fx=1limx ax2+bx2+1=1

Dividing the numerator and the denominator by x2:

limx a+bx21+1x2=1As x, 1x, 1x20a+01+0=1a=1a=1, b=1fx=x2+1x2+1=1f-2=1              Since fx is a constant function, its value does not depend on x.f2=1

Page No 29.39:

Question 22:

Show that limxx2+x+1-xlimxx2+1-x

Answer:

limx x2+x+1-x limx x2+1-xLHS:limx x2+x+1-xRationalising the numerator:limx x2+x+1-x x2+x+1+xx2+x+1+x=limx x2+x+1-x2x2+x+1+x=limx x+1x2+x+1+xDividing the numerator and the denominator by x:limx 1+1xx2+x+1x+1=limx1+1xx2+x+1x2+1=limx 1+1x1+1x+1x2+1When x, then 1x0.11+1=12RHS:limx x2+1-x                 from -

Rationalising the numerator:

limx x2+1-x x2+1+xx2+1+x=limx x2+1-x2x2+1+x=1=0limx x2+x+1-xlimx x2+1-x

Page No 29.39:

Question 23:

limx-4x2-7x+2x

Answer:

limx-4x2-7x+2x

Let x =-m
When n → – ∞, then m → ∞.

limm4m2+7m-2m =limm 4m2=7m-2m ×4m2+7m+2m4m2+7m+2m=limm4m2+7m-2m24m2+7m+2m=limm 4m2+7m-4m24m2+7m+2m

Dividing the numerator and the denominator by m:

limm 74m2+7mm2+2mm=limm74m2m2+7mm2+2=limm 74+7m+2As m , 1m0=74+2=74

Page No 29.39:

Question 24:

limx-x2-8x+x

Answer:

limx-x2-8x+x

Let x = –m

When x → –∞, then m → ∞.

limmm2+8m-m=limm m2+8m-m m2+8m+mm2+8m+m=limm m2+8m-m2m2+8m+m

Dividing the numerator and the denominator by m:

limm 8m2+8m+1m=limm8m2m2+8mm2+1=limm81+8m+1As m, 1m0=81+0+1=4

Page No 29.39:

Question 25:

Evaluate: limn14+24+34+...+n4n5-limn13+23+...+n3n5

Answer:

Consider the identity

k+15-k5=5k4+10k3+10k2+5k+1        .....(1)

Putting k = 1, 2, 3,..., n in (1) and then adding the equations, we have

n+15-1=5k=1nk4+10k=1nk3+10k=1nk2+5k=1nk+k=1n1n5+5n4+10n3+10n2+5n=5k=1nk4+10n2n+124+10nn+12n+16+5nn+12+n5k=1nk4=n5+5n4+10n3+10n2+4n-5n2n+122-5nn+12n+13-5nn+125k=1nk4=n5+5n42+5n33-n6

This expression on further simplification gives

k=1nk4=nn+12n+13n2+3n-130


limn14+24+34+...+n4n5-limn13+23+...+n3n5=limnnn+12n+13n2+3n-130n5-limnn2n+124n5=130limn1+1n2+1n3+3n-1n2-14limn1n1+1n2=130×1+0×2+0×3+0-0-14×0                                                      limn1n=limn1n2=...=0
=130×6-0=15

Page No 29.39:

Question 26:

Evaluate: limn1.2+2.3+3.4+...+nn+1n3

Answer:


limn1.2+2.3+3.4+...+nn+1n3=limnk=1nkk+1n3=limnk=1nk2+k=1nkn3=limnnn+12n+16+nn+12n3
=limnnn+12n+1+36n3=limn2nn+1n+26n3=13limn1+1n1+2n=13×1+0×1+0                  limn1n=0=13



Page No 29.49:

Question 1:

limx0sin 3x5x

Answer:

limx0sin 3x5x

=15limx0sin3x3x×3         limx0sin3x3x=1

15×1×3

35

Page No 29.49:

Question 2:

limx0sin x0x

Answer:

We know that x°=π180x.

limx0 sin x0x=limx0 sin π180xx=limx0 sin π180xπ180x×π180=π180×1=π180    

Page No 29.49:

Question 3:

limx0x2sin x2

Answer:

limx0x2sinx2

 When x0, then x20.

Let θ=x2

limθ0θsinθ

= 1

Page No 29.49:

Question 4:

limx0sin x cos x3x

Answer:

limx0sinx cosx3x

13limx0sinxx×cosx

13×1×cos0

13×1

= 13



Page No 29.50:

Question 5:

limx03 sinx-4 sin3xx

Answer:

limx03sinx-4sin3xx

limx0sin 3xx          sin3A=3sinA-4sin3A

limx0sin 3x3x×3

= 1 × 3

= 3

Page No 29.50:

Question 6:

limx0tan 8xsin 2x

Answer:

limx0tan 8xsin 2x

limx0tan 8x8x×8xsin 2x2x×2x              limx0tan 8x8x=1, limx0sin 2x2x=14

Page No 29.50:

Question 7:

limx0tan mxtan nx

Answer:

limx0tan mxtan nx

limx0tan mxmx×mxnx×nxtan nxlimx0tan mxmx×mxtan nxnx×nxmn                              limx0tan xx=1

Page No 29.50:

Question 8:

limx0sin 5xtan 3x

Answer:

limx0sin 5xtan 3x

limx0limx0sin 5x5x×5xtan 3x3x×3x53                                          limx0sin xx=1, limx0tan xx=1

Page No 29.50:

Question 9:

limx0sin xnxn

Answer:

limx0sin xnxn

It is of the form 00.

Let y=xnlimx0=limy0limy0sin yy1           limx0sin xx=1

Page No 29.50:

Question 10:

limx07x cos x-3 sin x4x+tan x

Answer:

limx07x cos x-3 sin x4x+tan x

It is of the form 00.
Dividing the numerator and the denominator by x:

limx07cosx-3sin xx4+tan xx7limx0cos x - 3limx0sin xx4+limx0tan xx7.1-3.14+145

Page No 29.50:

Question 11:

limx0cos ax-cos bxcos cx-cos dx

Answer:

limx0cos ax-cos bxcos cx-cos dx

It is of the form 00.

limx0-2sinax+bx2sinax-bx2-2sincx+dxa2sincx-dx2= limx0sinax+bx2ax+bx×ax+bx2×sinax-bx2ax-bx2×ax-bx2sincx+dx2cx+dx2×cx+dx2×sincx-dx2cx-dx2×cx-dx2=1×limx0ax+bx2ax-bx2cx+dx2cx-dx2= limx0x2 a+ba-bx2 c+dc-d= a2-b2c2-d2

Page No 29.50:

Question 12:

limx0tan2 3xx2

Answer:

limx0tan2 3xx2

= limx0tan 3x3x×tan 3x3x×9= 1 × 1 × 9                                  limx0tan xx=1= 9

Page No 29.50:

Question 13:

limx01-cos mxx2

Answer:

limx01-cos mxx2

= limx02sin2 mx2x2             1-cos A= 2sin2 A2= 2limx0sin mx2mx2×sin mx2mx2×m2×m2 = 2×m2×m2                                            limx0sin xx=1= m22

Page No 29.50:

Question 14:

limx03 sin 2x+2x3x+2 tan 3x

Answer:

limx03 sin 2x+2x3x+2 tan 3x

= limx03 sin 2xx+23+2 tan 3xx=3limx0sin 2x2x×2 + 23+2limx0 tan 3xx×3=6+23+6= 89

Page No 29.50:

Question 15:

limx0cos 3x-cos 7xx2

Answer:

limx0cos 3x-cos 7xx2
= limx0-2sin3x+7x2sin3x-7x2x2         cos C-cos D=-2sinC+D2sinC-D2= limx0-2sin 5x sin -2xx2= limx02sin 5x sin 2xx2                                  sin-θ = -sinθ= 2limx0sin 5x5x×sin 2x2x×5×2= 2 × 5 × 2=20

Page No 29.50:

Question 16:

limθ0sin 3θtan 2θ

Answer:

limθ0sin 3θtan 2θ

= limθ0sin 3θ3θ×3θtan 2θ2θ×2θ=32

Page No 29.50:

Question 17:

limx0sin x2 1-cos x2x6

Answer:

limx0sin x2 1-cos x2x6
= limx0sin x2×2sin2 x22x6          1-cos A=2sin2A2= 2limx0sin x2x2×sin x222×x22×sin x222×x22= 22×2= 12

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Question 18:

limx0sin2 4x2x4

Answer:

limx0sin2 4x2x4

= limx0sin4x2x2×sin4x2x2= limx0sin4x24x2×4×sin4x24x2×4= 4×4                           limx0sin xx=1= 16

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Question 19:

limx0x cos x+2 sin xx2+tan x

Answer:

limx0x cos x+2 sin xx2+tan x

Dividing the numerator and the denominator by x, we get:

 limx0cos x + 2sin xxx + tan xx= cos0+2×10+1                      limx0sin xx=1, limx0tan xx=1= 31

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Question 20:

limx02x-sin xtan x+ x

Answer:

limx02x-sin xtan x+ x

Dividing the numerator and the denominator by x, we get:

 limx02-sin xxtan xx+1= 2-11+1                          limx0tanxx=1, limx0sinxx=1= 12

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Question 21:

limx05 x cos x+3 sin x3x2+tan x

Answer:

limx05 x cos x+3 sin x3x2+tan x

Dividing the numerator and the denominator by x:

limx05cos x + 3sin xx3x + tan xx= 5cos0+33×0+1                          limx0sinxx=1, limx0tanxx=1= 5+30+1= 8

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Question 22:

limx0sin 3x -sin xsin x

Answer:

limx0sin 3x -sin xsin x

= limx02 cos3x+x2 sin3x-x2sinx                sin C-sin D = 2cosC+D2sinC-D2= limx02 cos2x.sinxsinx= 2cos0= 2

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Question 23:

limx0sin 5x - sin 3xsin x

Answer:

limx0sin 5x - sin 3xsin x

= limx02cos5x+3x2sin5x-3x2sin x         sinC-sinD=2cosC+D2sinC-D2= limx02cos4x. sin xsin x= 2 cos0= 2

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Question 24:

limx0cos 3x -cos 5xx2

Answer:

limx0cos 3x -cos 5xx2


= limx0-2sin3x+5x2 sin3x-5x2x2            cosC-cosD=-2sinC+D2sinC-D2= 2 limx02sin 4x sin xx2                             sin-θ=-sinθ= 2limx0sin 4x4x×4×sinxx= 2×4= 8.

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Question 25:

limx0tan 3x-2x3x-sin2x

Answer:

limx0tan 3x-2x3x-sin2x

Dividing the numerator and the denominator by x:

limx0tan 3xx-23-sin2 xx= limx0tan 3x3x×3 - 23-sinxx×sinx= 3-23-1×0                                limx0tan xx=1, limx0sinxx=1= 13

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Question 26:

limx0sin 2+x-sin 2-xx

Answer:

limx0sin 2+x-sin 2-xx

limx02cos 2+x+2-x2 sin2+x-2+x2xlimx02cos 2 sinxx2 cos 2                                       limx0sinxx=1

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Question 27:

limh0a+h2 sin a+h-a2 sin ah

Answer:

limh0a+h2 sin a+h-a2 sin ah

= limh0a2 sin a+h + h2sin a+h+2ah sina+h - a2 sin ah= limh0a2sina+h-sinah+h2sina+hh+2ah sina+hhDividing and multiplying the denominator by 2: limh0a22cosa+h+a2 sina+h-a22×h2+hsina+h+2a sina+h= limh0a2cos a+h+a2+h sin a+h+2a sin a+h= a2 cos a+0+2a sina= a2 cosa + 2a  sina

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Question 28:

limx0tan x -sin xsin 3x - 3 sin x

Answer:

limx0tan x -sin xsin 3x - 3 sin x

= limx0sinxcosx-sinx3sinx-4sin3x-3sinx= limx0sinx1-cosxcosx×-4sin3x= limx02sin2x2cosx×-4sin2x= limx02sinx2×sinx2cosx×-4 sinx×sinx= limx02sinx2x2×x2×sinx2x2×x2cosx×-4sinxx×x×sinxx×x= limx02×x2×x2cosx×-4×x×x= -18 cos0= -18

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Question 29:

limx0sec 5x -sec 3xsec 3x-sec x

Answer:

limx0sec 5x -sec 3xsec 3x-sec x

= limx01cos 5x1cos 3x1cos 3x-1cos x= limx0cos 3x - cos 5xcos 5x cos 3xcos x-cos 3xcos x cos 3x= limx0cos3x-cos5xcosxcos 5xcosx-cos3x= limx0-2sin3x+5x2sin3x-5x2×cosxcos5x-2sinx+3x2sinx-3x2         cosC-cosD=-2sinC+D2sinC-D2= limx0sin4x×sin-x×cosxcos5x×sin2x×sin-x= limx0sin4x4x×4xsin2x2x×2x×cosxcos5x= 42cos0cos0= 2

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Question 30:

limx01-cos 2xcos 2x-cos 8x

Answer:

limx01-cos 2xcos 2x-cos 8x

= limx02sin2 x2 sin2x+8x2sin8x-2x2                  cosC-cosD=2sinC+D2sinD-C2= limx02sin x × sin x2sin 5x×sin 3x= limx0sinx × sinxsin 5x × sin 3x                          sin-θ=-sinθ= limx0sinxx× x × sinxx × xsin 5x5x× 5x × sin 3x×3x3x= 115

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Question 31:

limx01-cos 2x+tan2xx sin x

Answer:

limx0 1-cos 2x+tan2 xx sin x=limx0 2 sin2 x+tan2 xx sin x            1-cos 2A=2 sin2 ADividing numerator & denominator by x2:limx0 2 sin2 xx2+tan2 xx2sin xx=212+121                     limx0 sin2 xx2=1, limx0 tan2 xx2=1=3

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Question 32:

limx0sin a+x+sin a-x-2 sin ax sin x

Answer:

limx0 sin a+x+sin a-x-2 sin ax sin x=limx0 2 sin a+x+a-x2 cos a+x-a+x2-2 sin ax sin x               sin C+sin D=2 sin C+D2cos C-D2=limx0 2 sin a cos x - 2 sin ax sin x=limx0  2 sin a cos x-1x sin x=limx0 2 sin a 1-2 sin2 x2-1x sin x=limx0 2 sin a -2 sin2 x2x sin x=-4 sin alimx0 1x sin xx2×sin x2x2×sinx2x2×14=-4 sin a×11×1×1×14=-sin a

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Question 33:

limx0x2 -tan 2xtan x

Answer:

limx0 x2-tan 2xtan xDividing the numerator and the denominator by x:limx0 x-tan 2xxtan xx=limx0 x-tan 2x2x×2tan xx=0-211=-2

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Question 34:

limx02-1+cos xx2

Answer:

limx0 2-1+cos xx2Rationalising the numerator:limx0 2-1+cos x 2+1+cos xx2×2+1+cos x=limx0 2-1+cos xx22+1+cos x=limx0 1-cos xx2 2+1+cos x=limx0 2 sin2 x2x2 2+1+cos x=2 limx0 sin2 x24×x22×12+1+cos x=2×14×12+1+cos 0=242+2=142

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Question 35:

limx0 x tan x1- cos x

Answer:

limx0 x tan x1-cos xDividing the numerator and the denominator by x2:limx0 x tan xx21-cos xx2=limx0 tan xx2 sin2 x2x2×x2×4=42=2

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Question 36:

limx0x2+1-cos xx sin x

Answer:

limx0 x2+1-cos xx sin x=limx0x2+2 sin2 x2x sin xDividing the numerator and the denominator by x2:limx0 1+2 sin2 x2x2sin xx=limx0 1+2 sin2 x2x22×4sin xx=1+241=321=32

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Question 37:

limx0sin 2x cos 3x-cos xx3

Answer:

limx0 sin 2x cos 3x-cos xx3=limx0 sin 2x×-2 sin3x+x2 sin3x-x2x3=-2 limx0 sin 2x2x×sin 2x2x×sin xx×2×2=-8

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Question 38:

limx02 sin x-sin 2x x3

Answer:

limx0 2 sin x°-sin 2x°x3=limx0 2 sin πx180-sin 2πx180x3=limx0 2 sin πx180-2 sin πx180×cosπx180x3=limx0 2 sin πx180 1-cos πx180x3=limx0 2 sin πx180×2 sin2 πx360x×x2=limx0 4 sin πx180× sin2 πx360πx180×πx360×πx360×π180×π3602=4 limx0  sin πx180πx180×sin πx360×sin πx360πx360×πx360×π3180×3602=4×1×1×1×π3180×360×360=π1803

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Question 39:

limx0x3 cot x1-cos x

Answer:

limx0 x3 cot x1-cos x=limx0 x3tan x 1-cos x=limx0 x3tan x×2 sin2 x2=limx0 xtan x×x22 sin2 x2=limx0 xtan x×x24×42×sin2 x2=limx0 xtan x×x2sin x22×42=1×1×42=2

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Question 40:

limx0x tan x1-cos 2x

Answer:

limx0 x tan x1-cos 2x=limx0 x tan x2 sin2 xDividing the numerator and the denominator by x2:limx0 x tan xx22sin2 xx2=limx0 tan xx2sin xx2=12

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Question 41:

limx0sin 3+x-sin 3-xx

Answer:

limx0 sin 3+x-sin 3-xx=limx0 2 cos 3+x+3-x2 sin 3+x-3+x2x      sin C-sin D=2 cos C+D2 sin C-D2=limx0 2 cos 3 sin xx=2 cos 3

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Question 42:

limx0 cos 2x-1cos x-1

Answer:

limx0 cos 2x-1cos x-1=limx0 1-2 sin2 x-11-2 sin2 x2-1=limx0 sin2 xsin2 x2=limx0 sin2 xx2×x2sin2 x2x24×x24=limx0 sin xx2×1sin x2x22×4=4



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Question 43:

limx03 sin2 x-2 sin x23x2

Answer:

limx0 3 sin2 x-2 sin x23x2=limx0 3 sin2 x3x2-2 sin x23x2=1-23=13

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Question 44:

limx01+sin x-1-sin xx

Answer:


limx0 1+sin x-1-sin xx=limx0 1+sin x-1-sin x 1+sin x+1-sin xx1+sin x+1-sin x=limx0 1+sin x-1-sin xx1+sin x+1-sin x=limx0 2 sin xx1+ sin x+1-sin x=21+0+1-0=22=1

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Question 45:

limx01-cos 4xx2

Answer:

limx0 1-cos 4xx2=limx0 2 sin2 2xx2=limx0 2sin 2xx×sin 2xx=limx0 2×sin 2x2x×sin2x2x×4=2×1×1×4=8

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Question 46:

limx0x cos x+sin xx2+tan x

Answer:

limx0 x cos x+sin xx2+tan xDividing the numerator and the denominator by x:limx0 cos x+sin xxx+tan xx=cos 0+10+1=2

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Question 47:

limx01-cos 2x3 tan2 x

Answer:

limx0 1-cos 2x3 tan2 x=limx0 2 sin2 x3 tan2 x=limx0 23×sin2 xsin2 x×cos2 x=23 cos2 0=23

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Question 48:

limθ01-cos 4θ1-cos 6θ

Answer:

limθ0 1-cos 4θ1-cos 6θ=limθ0 2 sin2 2θ2 sin2 3θ           1-cos A=2 sin2 A2=limθ0 sin2 2θ2θ2×2θ2sin2 3θ3θ2×3θ2=limθ0 sin 2θ2θ2×3θsin 3θ2×49=49                         limX0 sin xx=1

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Question 49:

limx0ax+x cos xb sin x

Answer:

limx0 ax+x cos xb sin xDividing the numerator and the denominator by x:limx0 a+cos xbsin xx                             limx0 sin xx=1=a+cos 0 b=a+1b

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Question 50:

limθ0sin 4θtan 3θ

Answer:

limθ0 sin 4θtan 3θ=limθ0 sin 4θ4θ×4θtan 3θ3θ×3θ=43

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Question 51:

Evaluate the following limits:

limx02sinx-sin2xx3

Answer:


limx02sinx-sin2xx3=limx02sinx-2sinxcosxx3=limx02sinx1-cosxx3=limx02sinx×2sin2x2x3
=limx0sinx×sin2x2x×x24=limx0sinxx×limx0sinx2x22=1×1                                                                            limθ0sinθθ=1=1

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Question 52:

limx01-cos 5x1-cos 6x

Answer:

limx0 1-cos 5x1-cos 6x=limx0 2 sin2 5x22 sin2 3x=limx0 sin2 5x25x22×5x22sin2 3x3x2×3x2=limx0 sin 5x25x22254x2sin 3x3x2×9x2=259×4=2536

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Question 53:

limx0cosec x-cot xx

Answer:

limx0 cosec x-cot xx=limx0 1sin x-cos xsin xx=limx0 1-cos xx sin x=limx0 2 sin2 x2x sin x=limx0 2 sin2 x2x22×x24x sin xx×x×x2=2×14=12

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Question 54:

limx0sin 3x+ 7x4x+sin 2x

Answer:

limx0 sin 3x+7x4x+sin 2x=limx0 sin 3x3x×3x + 7x4x + sin 2x2x×2x=limx0 sin 3x3x×3+7x4+sin 2x2x×2x=3+74+2=106=53

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Question 55:

limx05x+4 sin 3x4 sin 2x+7x

Answer:

limx0 5x+4 sin 3x4 sin 2x+7x=limx0 5x+4×sin 3x3x×3x4 sin 2x2x×2x+7x=limx0 5+4 sin 3x×33xx4sin 2x2x×2+7x=5+4×34×2+7                               limx0 sin 3x3x=1=1715

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Question 56:

limx03 sin x-sin 3xx3

Answer:

limx0 3 sin x-sin 3xx3=limx0 3 sin x-3 sin x-4 sin3 xx3=limx0 4 sin3 xx3=limx0 4sin xx3=4×1=4

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Question 57:

limx0tan 2x-sin 2xx3

Answer:

limx0 tan 2x-sin 2xx3=limx0 sin 2xcos 2x-sin 2xx3=limx0 sin 2x 1-cos 2xcoS 2x×x3=limx0 sin 2x×2 sin2 xcos 2x×x3=limx0 sin 2x2x×2cos 2x×2sin xx2=2×2cos 0=4

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Question 58:

limx0sin ax+bxax+sin bx

Answer:

limx0 sin ax+bxax+sin bx=limx0 sin axax×ax+bxax+sin bxbx×bx=limx0 sin axax×a+bxa+sin bxbx×bx=1×a+ba+b=1

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Question 59:

limx0 cosec x-cot x

Answer:

limx0 cosec x-cot x=limx0 1sin x-cos xsin x=limx0 1-cos xsin x                                 1-cos A=2 sin2 A2=limx0  2 sin2 x22 sin x2 cos x2=limx0 tan x2=0

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Question 60:

Evaluate the following limits:

limx0sinα+βx+sinα-βx+sin2αxcos2βx-cos2αx

Answer:


limx0sinα+βx+sinα-βx+sin2αxcos2βx-cos2αx=limx02sinα+βx+α-βx2cosα+βx-α-βx2+2sinαx cosαx1-sin2βx-1-sin2αx=limx02sinαx cosβx+2sinαx cosαxsin2αx-sin2βx=limx02sinαxcosβx+cosαxsin2αx-sin2βx
=limx02αx×sinαxαx×cosβx+cosαxα2x2×sin2αxα2x2-β2x2×sin2βxβ2x2=2α×limx0sinαxαx×limx0cosβx+cosαxα2×limx0sinαxαx2-β2×limx0sinβxβx2×limx0xx2=2α×1×1+1α2×1-β2×1×limx01x=4αα2-β2×=

Disclaimer: There is some mistake or some misprinting in the question.

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Question 61:

Evaluate the following limits:

limx0cosax-cosbxcoscx-1

Answer:


limx0cosax-cosbxcoscx-1=limx0-2sinax+bx2sinax-bx2-2sin2cx2                               1-cos2θ=2sin2θ=limx0a+b2x×sina+b2xa+b2x ×a-b2x×sina-b2xa-b2xc2x24×sin2cx2c2x24=a+ba-bc2×limx0sina+b2xa+b2x×limx0sina-b2xa-b2xlimx0sincx2cx22
=a2-b2c2×1×11                                                                          limθ0sinθθ=1=a2-b2c2

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Question 62:

Evaluate the following limits:

limh0a+h2sina+h-a2sinah

Answer:


limh0a+h2sina+h-a2sinah=limh0a2+2ah+h2sina+h-a2sinah=limh02ah+h2sina+h+a2sina+h-a2sinah=limh02ah+h2sina+hh+limh0a2sina+h-a2sinah=limh02a+hsina+h+a2limh0sina+h-sinah
=2a+0sina+0+a2limh02sinh2cos2a+h2h=2asina+a2×limh0sinh2h2×limh0cos2a+h2=2asina+a2×1×cos2a+02                                                   limθ0sinθθ=1=2asina+a2cosa             

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Question 63:

If limx0 kx cosec x=limx0 x cosec kx, find k.

Answer:

limx0 kx.cosec x=limx0 x cosec kxlimx0 kxsin x=limx0 xsin kxklimx0 xsin x=limx0 kxsin kx×1kk=1kk2=1k=±1



Page No 29.62:

Question 1:

limxπ sin xπ-x

Answer:

limxπ sin xπ-x=limh0 sin π-hπ-π-h                       limxafx=limh0fa-h=limh0 sin hh                              sin π-0=sin 01

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Question 2:

limxπ2 sin 2xcos x

Answer:

limxπ2 sin 2xcos x                            sin 2x= 2 sin x cos x=limxπ2 2 sin x cos xcos x=limxπ2  2 sin x2

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Question 3:

limxπ2cos2x1-sin x

Answer:

limxπ2 cos2 x1- sin x=limxπ2 1-sin2 x1-sin x                                         cos2 x=1-sin2 x    a2-b2=a-b a+b=limxπ2 1-sin x 1+sin x1-sin x=limxπ2 1+sin x1+1=2

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Question 4:

Evaluate the following limits:

limxπ31-cos6x2π3-x

Answer:


limxπ31-cos6x2π3-x=limxπ32sin23x2π3-x                                      1-cos2θ=2sin2θ=limxπ32sin3x2π3-x=limxπ3sin3xπ3-x
=limh0sin3π3+hπ3-π3+h                           Put x=π3+h=limh0sinπ+3h-h=limh0-sin3h-h                                        sinπ+θ=-sinθ=3×limh0sin3h3h=3×1                                                      limθ0sinθθ=1=3

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Question 5:

limxa cos x-cos ax-a

Answer:

limxa cos x-cos ax-a=limxa -2 sin x+a2 sin x-a22x-a2                cos A-cos B -2 sin A-B2 sin A+B2=limxa -sin x+a2                                          limθa sinθ-aθ-a=1=-sin 2a2-sin a

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Question 6:

limxπ4 1-tan xx-π4

Answer:

limxπ4 1-tan xx-π4=limh0 1-tan π4+hπ4+h-π4=limh0 1-1+tan h1-tan hh               tan π4+0=1+tan θ1-tan θ=limh0 1-tan h-1-tan h1-tan h h=limh0 -2 tan hh1-tan h                      limh0 tan hh=1-21-0=-2

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Question 7:

limxπ2 1-sin xπ2-x2

Answer:

limxπ21-sin xπ2-x2=limh0 1-sin π2-hπ2-π2-h2=limh0 1-cos hh2=limh0 2 sin2 h24h24                        limh0 sin hh=1=12 limh0 sin h2h2212

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Question 8:

limxπ3 3-tan xπ-3x

Answer:

limxπ3 3-tan xπ-3x=limh0 3-tan π3-hπ-3π3-h=limh03 -tan π3-tan h1+tan π3 tan hπ-3π3-h=limh0  3-3-tan h1+3 tan h3h=limh0 3+3 tan h-3+tan h1+3 tan h 3h=limh0 4 tan h3h 1+3tan h=43

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Question 9:

limxπ2 cot xπ2-x

Answer:

limxπ2 cot xπ2-x=limh0 cot π2-hπ2-π2-hlimx0 tan hh=1

Page No 29.62:

Question 10:

limxπ4 cos x-sin xx-π4

Answer:

limxπ4 cos x-sin xx-π4Rationalising the numerator:=limxπ4 cos x-sin xx-π4 cos x+sin xcos x+sin x=limxπ4 cos x-sin xx-π4 cos x+sin x=limh0 cos π4+h-sin π4+hπ4+h-π4 cos π4+h+sin π4+h=limh0 cos π4 cos h - sin π4 sin h - sin π4 cos h - cos π4 sin hhcos π4+h+sin π4+h=limh0 -2 sin hhcos π4+h+sin π4+h-2×121214=-1214=-2-14

Page No 29.62:

Question 11:

limxπ2 2-sin x-1π2-x2

Answer:

limxπ2 2-sin x-1π2-x2=limh0 2-sin π2-h-1π2-π2-h2=limh0 2-cos h-1h2Dividing the numerator and the denominator by 2-cos h+1:limh0 2-cos h-1 2-cos h+12-cos h+1 h2=limh0 2-cos h-1h2 2-cos h+1=limh0 1-cos hh22-cos h+1=limh0 2 sin2 h24h24 2-cos h+1=12limh0 sin h2h22×limh0 1 2-cos h+1=12 2-1+1=14

Page No 29.62:

Question 12:

limxπ4 2 -cos x-sin xπ4-x2

Answer:

limxπ4 2-cos x-sin xπ4-x2Dividing the numerator and the denominator by 2:limxπ41-12cos x+12sin x12 π4-x2=limxπ41-cos π4cos x+sin π4 sin x12 π4-x2=limxπ4 21-cos π4-xπ4-x2=limxπ4 2 2 sin2 π4-x2π4-x2                    1-cos θ=2 sin 2θ2=22limxπ4  sin2 π4-x24π4-x22=224=12

Page No 29.62:

Question 13:

limxπ8 cot 4x-cos 4xπ-8x3

Answer:

limxπ8 cot 4x-cos 4xπ-8x3=limh0 cot 4π8-h-cos 4π8-hπ-8π8-h3=limh0 tan 4h-sin 4h8 h3=limh0 sin 4h-cos 4h sin 4h512 cos 4hh3=limh0sin 4h 1-cos 4hcos 4h 512 h3=limh0 tan 4h4h×2 sin2 2h32×4h2=116limh0tan 4h4h×limh0sin 2h2h2=116×1×1 = 116

Page No 29.62:

Question 14:

limxa cos x-cos ax-a

Answer:

limxa cos x-cos ax-a=limxa -2 sin x+a2 sin x-a2x-aDividing the numerator and the denominator by x+a:-2 limxa sin x+a2sin x-a2 2x-a2x+a=-2limxa  sin x+a2×sin x-a22x-a2x+a-2 sin 2a2 a+a2=-2a sin a

Page No 29.62:

Question 15:

limxπ 5+cos x-2π-x2

Answer:

limxπ 5+cos x-2π-x2=limh0 5+cos π-h-2π-π-h2=limh0 5-cos h-2h2Rationalising the numerator, we get:limh0 5-cos h-2 5-cos h+2h2 5-cos h+2=limh0 5-cos h-4h2 5-cos h+2=limh0 1-cos hh2 5-cos h+2=limh0 2 sin2 h24h24 5-cos h+2=12 5-1+2=124=18

Page No 29.62:

Question 16:

limxa cos x-cos ax-a

Answer:

limxa cos x-cos ax-a=limxa -2 sin x+a2 sin x-a22x-a2 x+a=limxa  -sin x+a2 x+a ×sin x-a2x-a2=-1 sin a2a×1=-sin a2a

Page No 29.62:

Question 17:

limxa sin x-sin ax-a

Answer:

limxa sin x-sin ax-a=limxa 2 cos x+a2 sin x-a2x+a x-a=limxa 2 cos x+a2x+a×sin x-a22x-a2=1a+a cos 2a212a cos a

Page No 29.62:

Question 18:

limx1 1-x2sin 2π x

Answer:

limx1 1-x2sin 2πx=limh0 1-1-h2sin 2π1-h=limh0 2h-h2- sin 2π h=limh0 -2-hsin 2π hh=limh0 h-22π sin 2π h2π h=0-22π×1=-1π

Page No 29.62:

Question 19:

limxπ4 fx-fπ4x-π4, where f(x) = sin 2x

Answer:

limxπ4 fx-fπ4x-π4=limh0 fπ4+h-fπ4π4+h-π4It is given that  fx=sin 2x.limh0 sin π2+2h-sin π2h=limh0 cos 2h-1h=limh0 -2sin2 hh×h hlimh0 -2h=0

Page No 29.62:

Question 20:

limx1 1+cos πx1-x2

Answer:

limx1 1+cos π x1-x2=limh0 1+cos π 1-h1-1-h2=limh0 1-cos πhh2=limh0 2 sin2 πh24π2π2h24=limh0 sin2 πh22π2 πh22π22

Page No 29.62:

Question 21:

limx1 1-x2sin π x

Answer:

limx1 1-x2sin πxlimh0 1-1-h2sin π1-hlimh0 2h-h2sin πh=limh0 h2-hsin π h=limh0 2-hπ×sin πhπh2-0π=2π

Page No 29.62:

Question 22:

limxπ4 1-sin 2x1+cos 4x

Answer:

limxπ4 1-sin 2x1+cos 4x=limh0 1-sin 2π4-h1+cos 4π4-h=limh0 1-cos 2h1-cos 4h=limh0 2 sin2 h2 sin2 2hlimh0 sin2 hh24 sin2 2h4h214

Page No 29.62:

Question 23:

limx1 1-1xsin π x-1

Answer:

limx11-1xsinπx-1=limx1x-1xsinπx-1

Let y = x – 1
If x → 1, then y → 0.

=limy0yy+1sinπy=limy01πy+1 sin πyπy=1π0+1×1=1π

Page No 29.62:

Question 24:

limnn sin π4 n cos π4 n

Answer:

limnn sin π4ncos π4n=limn n sin π4n×limn cos π4n=limn n sin π4nπ4n×π4n×1=π4limn sin π4nπ4nLet y=π4nIf n, then y0.=π4limy0sin yy=π4

Page No 29.62:

Question 25:

limn2n-1 sin a2n

Answer:

limn 2n-1 sin a2n=limn 2n2×sin a2na2n× a2nLet y=a2nIf n, then y0.=limy0 a2×sin yya2

Page No 29.62:

Question 26:

limn sin a2nsin b2n

Answer:

limn sin a2nsin b2n=limn a2n sin a2na2n×b2n× sin b2nb2nLet: y=a2n z=b2nIf n, then y0 and z0.=yzlimy0sin yy×1limz0sin zz=yz×1×11=a2n×1b2n×1                  =ab

Page No 29.62:

Question 27:

limx-1 x2-x-2x2+x+sin x+1

Answer:

limx-1 x2-x-2x2+x+sin x+1=limx-1 x-2 x+1xx+1+sin x+1Let y=x+1If x-1, then y0.=limy0  y-3yy-1y+sin yDividing the numerator and the denominator by y:=limy0 y-3y-1 +sin yy=0-30-1+1-30=

Page No 29.62:

Question 28:

limx2 x2-x-2x2-2x+sin x-2

Answer:

limx2 x2-x-2x2-2x+sin x-2=limx2 x-2 x+1xx-2+sin x-2Let y=x-2 x2 y0=limy0 yy+3y+2y+sin yDividing the numerator and the denominator by y:=limy0 y+3y+2+sin yy=32+1=33=1



Page No 29.63:

Question 29:

limx1 1-x tan πx2

Answer:

limx1 1-x tanπx2=limh0 1-1-h tan π2 1-h=limh0 h tan π2-πh2=limh0 h cot πh2=limh0 htan πh2=limh0 1tanπh2×π2πh2=1π2                             limh0 tan hh=1=2π

Page No 29.63:

Question 30:

limxπ4 1-tan x1-2 sin x

Answer:

limxπ4 1-tan x1-2 sin xIt is of 00 form.

Rationalising the denominator, we get:

limxπ4 1-tan x 1+2 sin x1-2 sin x 1+2 sin x=limxπ4 1-tan x 1+2 sin x1-2 sin2 x=limxπ4 1-sin xcos x 1+2 sin xcos 2x=limxπ4 cos x-sin x 1+2 sin xcos x cos 2x =limxπ4 cos x-sin x 1+2 sin xcos x·cos2 x-sin2 x=limxπ4 cos x-sin x 1+2 sin xcos x cos x-sin x cos x+sin x=1+2×1212 12+12=212×2=2

Page No 29.63:

Question 31:

limxπ 2+cos x-1π-x2

Answer:

limxπ 2+cos x-1π-x2=limh0 2+cos π-h-1π-π-h2=limh0 2-cos h-1h2×2-cos h+12-cos h+1=limh0 2-cos h-1h22-cos h+1=limh0 1-cos hh2 2-cos h+1=limh0 2 sin2 h24×h24× 2-cos h+1=1×122 2-1+1=14

Page No 29.63:

Question 32:

limxπ 1+cos xtan2x

Answer:

limxπ 1+cos xtan2 x=limh0 1+cos π+htan2 π+h=limh0 1-cos htan2 h=limh0 2 sin2 h2tan2 h=limh0 2 sin2 h24h24×tan2 hh2=12 limh0sin h2h22limh0tan hh2=12×1212=12

Page No 29.63:

Question 33:

limxπ2 π2-x tan x

Answer:

limx1 1-1xsin π x-1=limx1 x-1x sin πx-1Let y=x-1 x1 y0=limy0 yy+1 sin π y=limy0 1πy+1×sin πyπy=1π0+1×1=1π

Page No 29.63:

Question 34:

limxπ6 cot2 x-3cosec x-2

Answer:

limxπ6 cot2 x-3cosec x-2=limxπ6 cosec2 x-1-3cosec x-2=limxπ6 cosec2 x-4cosec x-2=limxπ6 cosec x-2 cosec x+2cosec x-2=cosecπ6+2=4

Page No 29.63:

Question 35:

limxπ4 2-cos x-sin x4x-π2

Answer:

limxπ4 2-cos x-sin x4x-π2=limh0 2-cos π4+h+sin π4+h4π4+h-π2=limh0 2-cos π4 cos h - sin π4 sin h + sin π4 cos h + cos π4 sin h4h2=limh0 2-2cos h4h2=limh0 2 1-cos h16 h2=limh0 22 sin2 h264×h24=232×12=1162

Page No 29.63:

Question 36:

limxπ2 π2-x sin x-2 cos xπ2-x+cot x

Answer:

limxπ2 π2-x sin x-2 cos xπ2-x+cot x=limh0 π2-π2-h sin π2-h-2 cos π2-hπ2-π2-h+cot π2-h    Plugging x=π2-h=limh0 h cos h-2 sin hh+tan hDividing the numerator and the denominator by h:limh0 cos h- 2 sin hh1+tan hh1-211+1=-12

Page No 29.63:

Question 37:

limxπ4 cos x-sin xπ4-x cos x + sin x

Answer:

limxπ4 cos x-sin xπ4-x cos x+sin xDividing the numerator and the denominator by2:limxπ412 cos x-12 sin xπ4-x cos x+sin x2 =limxπ4 2 sin π4 cos x-cos π4 sin xπ4-x cos x+sin x=limxπ4 2 sin π4-xπ4-x cos x+sin x=limxπ4 2sin x+cos x×limxπ4 sin π4-xπ4-x212+12×1=1

Page No 29.63:

Question 38:

Evaluate the following limits:

limxπ1-sinx2cosx2cosx4-sinx4                                         [NCERT EXEMPLAR]

Answer:


limxπ1-sinx2cosx2cosx4-sinx4

Put x=π+h

When xπ, h0

limxπ1-sinx2cosx2cosx4-sinx4=limh01-sinπ+h2cosπ+h2cosπ+h4-sinπ+h4=limh01-sinπ2+h2cosπ2+h2cosπ4+h4-sinπ4+h4=limh01-cosh2-sinh2cosπ4cosh4-sinπ4sinh4-sinπ4cosh4+cosπ4sinh4
=limh02sin2h4-2sinh4cosh412cosh4-12sinh4-12cosh4-12sinh4=limh0sinh4-cosh4×-2sinh4=12×1limh0cosh4=12×1                               limθ0cosθ=1=12                                   



Page No 29.65:

Question 1:

limxπ 1+cos xtan2x

Answer:

limxπ 1+cos xtan2 x                             00form=limxπ 1+cos xsin2 x×cos2 x=limxπ 1+cos x1-cos2 x×cos2 x=limxπ 1+cos x1-cos x 1+cos x×cos2 x=limxπ cos2 x1-cos x=cos2 π1-cos π=-121--1=12

Page No 29.65:

Question 2:

limxπ4 cosec2 x-2cot x-1

Answer:

limxπ4 cosec2 x-2cot x-1=limxπ4 1+cot2 x-2cot x-1=limxπ4 cot2 x-1cot x-1=limxπ4 cot x-1 cot x+1cot x-1=cot π4+1=2

Page No 29.65:

Question 3:

limxπ6 cot2 x-3cosec x-2

Answer:

limxπ6 cot2 x-3cosec x-2=limxπ6 cosec2 x-1-3cosec x-2=limxπ6cosec2 x-4cosec x-2=limxπ6 cosec x-2 cosec x+2cosec x-2=cosecπ6+2=2+2=4

Page No 29.65:

Question 4:

limxπ4 2-cosec2x1-cot x

Answer:

limxπ4 2-cosec2 x1-cot x=limxπ4 2-1+cot2 x1-cot x=limxπ4 1-cot2 x1-cot x=limxπ4 1-cot x 1+cot x1-cot x=1+cot π4=1+1=2

Page No 29.65:

Question 5:

limxπ 2+cos x-1π-x2

Answer:

limxπ 2+cos x-1π-x2Rationalising the numerator, we get:limxπ 2+cos x-1×2+cos x+1π-x2 2+cos x+1=limxπ 2+cos x-1π-x2 2+cos x+1=limxπ 1+cos xπ-x2 2+cos x+1

Let x = π - h
when x → π, then h → 0

=limh0 1+cos π-hπ-π-h2 2+cos π-h+1=limh0 1-cos hh2 2-cos h+1                      cos π-θ=-cos θ=limh0 2 sin2 h24×h242-cos h+1=12limh0 sin h2h22×12-cos h+1=12×1×12-cos 0+1=12×11+1=12×2=14

Page No 29.65:

Question 6:

limx3π2 1+cosec3 xcot2 x

Answer:

limx3π3 1+cosec3 xcot2 x=limx3π21+cosec x 12+cosec2 x-cosec xcosec2 x-1=limx3π2 1+cosec x 1+cosec2 x-cosec xcosec x-1 cosec x+1=1+cosec2 3π2-cosec 3π2cosec 3π2-1=1+1+1-1-1=-32



Page No 29.71:

Question 1:

limx0 5x-14+x-2

Answer:

limx0 5x-14+x-2

Rationalising the denominator, we get:

=limx0 5x-1 4+x+24+x-2 4+x+2=limx0 5x-1 4+x+24+x-4=limx0 5x-1x 4+x+2                           limx0 ax-1x=log a=log 5×4+0+2=4 log 5

Page No 29.71:

Question 2:

limx0 log 1+x3x-1

Answer:

limx0 log 1+x3x-1

Dividing the numerator and the denominator by x:

limx0 log 1+xx·3x-1x                        limx0 log 1+xx=1 limx0 ax-1x=log a=1log 3

Page No 29.71:

Question 3:

limx0 ax+a-x-2x2

Answer:

limx0 ax+a-x-2x2=limx0 ax22+a-x22-2ax2·a-x2x2=limx0 ax2-a-x22x2=limx0 ax2-1ax22x2=limx0 ax-12x2×1ax22=limx0 ax-1x2×1ax=log a2a0=log a2

Page No 29.71:

Question 4:

limx0 amx-1bnx-1, n0

Answer:

limx0 amx-1bnx-1=limx0 amx-1mx×mxbnx-1nx×nx=loge aloge b×mn=mn log alog b

Page No 29.71:

Question 5:

limx0 ax+bx-2x

Answer:

limx0 ax+bx-2x=limx0 ax-1+bx-1x=limx0 ax-1x+limx0 bx-1x=log a+log b=log ab

Page No 29.71:

Question 6:

limx0 9x-2.6x+4xx2

Answer:

limx0 9x-2.6x+4xx2=limx0 3x2-2·3x·2x+2x2x2=limx0 3x-2x2x2=limx0 3x-2x2x2×2x2x2=limx0 32x-1x2×22x=log 322×20=log 322

Page No 29.71:

Question 7:

limx0 8x-4x-2x+1x2

Answer:

limx0 8x-4x-2x+1x2=limx0 2x3-2x2-2x+1x2=limx0 2x2 2x-1-12x-1x2=limx0 22x-1 2x-1x2=limx0 222x-12x×2x-1x=2 log 2×log 2=log 22×log 2=log 4×log 2

Page No 29.71:

Question 8:

limx0 amx-bnxx

Answer:

limx0 amx-bnxx=limx0 amx-1-bnx+1x=limx0 amx-1-bnx-1x=limx0 amx-1mx×m-bnx-1nx×n=m log a-n log b=log am-log bn=log ambn

Page No 29.71:

Question 9:

limx0 ax+bx+cx-3x

Answer:

limx0 an+bn+cn-3x=limx0 an-1x+bn-1x+cn-1x=log a+log b+log c=log abc

Page No 29.71:

Question 10:

limx2 x-2logax-1

Answer:

limx2 x-2loga x-1

Let x = 2 + h

 x → 2
h → 0

=limh0 2+h-2log 2+h-1log a=log a limh0 hlog 1+h=log a ×1

Page No 29.71:

Question 11:

limx0 5x+3x+2x-3x

Answer:

limx0 5x+3x+2x-3x=limx0 5x-1x+3x-1x+2x-1x=log 5+log 3+log 2=log 5×3×2=log 30

Page No 29.71:

Question 12:

limx a1/x-1x

Answer:

limx a1x-1xLet y=1x x y0limy0 ay-1y=log a

Page No 29.71:

Question 13:

limx0 amx-bnxsin kx

Answer:

limx0 amx-bnxsin kx=limx0 amx-1-bnx-1sin kxDividing the numerator and the denomiantor by x:=limx0 mamx-1mx-nbnx-1nxk×sin kxkx=m log a-n log bk×1=1k log am-log bn=1k log ambn

Page No 29.71:

Question 14:

limx0 ax+bx-cx-dxx

Answer:

limx0 an+bn-cn-dnxlimx0 an+bn-2-cn-dn+2x=limx0 an-1+bn-1-cn-1-dn-1x=limx0 an-1x+bn-1x-cn-1x-dn-1x=log a +log b-log c-log d=log a +log b-log c+log d=log ab-log cd=log abcd

Page No 29.71:

Question 15:

limx0 ex-1+sin xx

Answer:

limx0 ex-1+sin xx=limx0 ex-1x+sin xx=1+1=2

Page No 29.71:

Question 16:

limx0 sin 2xex-1

Answer:

limx0 sin 2xex-1

Dividing the numerator and the denominator by x:

=limx0 sin 2xx×ex-1x=limx0 sin 2x2x×2ex-1x=1×21=2

Page No 29.71:

Question 17:

limx0 esin x-1x

Answer:

limx0 esin x-1x=limx0 esin x-1sin x×sin xx

 x → 0
∴ sin x → 0

Let y=sin x

x → 0
y → 0

limy0 ey-1y×limx0 sin xx=1×1

Page No 29.71:

Question 18:

limx0 e2x-exsin 2x

Answer:

limx0 e2x-exsin 2x=limx0 ex ex-1sin 2x=limx0 ex ex-1x×2xsin 2x×12=e0×1×11×12=12

Page No 29.71:

Question 19:

limxa log x- log ax-a

Answer:

limxa log x-log ax-a=limxa log xaaxa-1=limxa log 1+xa-1axa-1xa xa1xa-10Let y=xa-1xa y0=limy0 log 1+ya×y=1a×1=1a

Page No 29.71:

Question 20:

limx0 log a+x-log a-xx

Answer:

limx0 log a+x-log a-xx=limx0 log a+xa-xx=limx0 log 1+a+xa-x-1x=limx0 log 1+a+x-a+xa-xx=limx0 log 1+2xa-x2xa-x×a-x2x0 2xa-x0Let y=2xa-x=limy0 log 1+yy ×limx0 1a-x2=1×2a

Page No 29.71:

Question 21:

limx0 log 2+x+log 0.5x

Answer:

limx0 log 2+x+log 0.5x=limx0 log 2+x×0.5x=limx0 log 1+x2x2×2=12×1=12

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Question 22:

limx0 log a+x-log ax

Answer:

limx0 log a+x-log ax=limx0 log a+xax=limx0 log 1+xaxa×a=1a×1=1a

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Question 23:

limx0 log 3+x-log 3-xx

Answer:

limx0 log 3+x-log 3-xx=limx0 log 3+x3-xx=limx0 log 1+3+x3-x-1x=limx0 log 1+3+x-3+x3-xx=limx0 log 1+2x3-xx=limx0 log 1+2x3-x2x3-x×3-x2=1×23-0=23

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Question 24:

limx0 8x-2xx

Answer:

limx0 8x-2xx=limx0 8x-1x-2x-1x=log 8-log 2=log 82=log 4

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Question 25:

limx0 x2x-11-cos x

Answer:

limx0 x2x-11-cos x

Dividing the numerator and the denominator by x2:

=limx0 2x-1x×x21-cos&#