# lim x tends to zero 1-cos4x / x2 ?

8

• -4

1-cos4x / x2 = 2sin2x// 2 / x2

Solving it we get 8.

• -1

lim x tends to 0   1-cos4x /x=r lim x tends to 0r 1-1+2sin2(2x)/x[since cos2x=1-sin2 x]

=>  lim x tends to 0  2sin2(2x)/x2

Multiplying 4  in numerator and denominator,

=>lim x tends to 0r 8sin2(2x)/(2x)2

=>8

Therefore ,  lim x tends to 0 1-cos4x/x2 = b

• 1

lim x-0

lim x-0=

lim x-0=cos 6x-1/3tan

2

x=

l

• -5
1-cos4x / x^2
=2 sin4x/2 / x^2
=2 sin2x / x^2
=2 sin2x × 2
= 4 / x
• -5
limx-->0 1-cos4x/x^2 (form 0/0)
so L.H rule is applicable. Applying we get
=4limx-->0sin4x/2x
=8 lim4x-->0sin4x/4x
=8 (using lim x-->0 sinx/x=1)
• 1
lim x-0 1-cos4x/x*2
lim x-0 (1-(1-2sin*2 2x)/x*2
lim x-0 (2sin*2 2x)/x*2
2 lim x-0(sin*2 2x)/x*2
multipying by 2
4 lim x-0 (sin*2 2x)/x*2
expand sin*2 2x/2 to sin 2x/2x *sin 2x/x
mltiply by 2 again since sin 2x/2x =1
8 lim x-0 sin 2x/2x
​8*1 = 8

• 5 • -6
Lim x tends to 0 ( tan2x-2sinx)/x³
• 2
how can i solve this... • 0 • 3
simplify 95 by may 6 - 4 3 18 + 27 by 12
• 0 • 0
a?(b+c)
• 0
solution one
• 0 • 0
What are the values of.
• 0