Lim (x approaches1) [(ln(1+x) - ln2)(3 4x-1 - 3x)] / [(7+x)1/3 - (1 + 3x)1/2]sin(x-1) - Maths - Limits and Derivatives

Question

Lim (x approaches1) [(ln(1+x) - ln2)(3 4x-1 - 3x)] /
[(7+x)1/3 - (1 + 3x)1/2]sin(x-1)

Ajanta Trivedi · Accepted Answer

the given limit is:
limx→1{log(1+x)-log2}{3*4x-1-3x}{(7+x)1/3-(1+3x)1/2}sinπx
=limh→0{log(2+h)-log2}{3*4h-3(1+h)}{(8+h)1/3-(4+3h)1/2}sin(π+πh)=limh→0log(1+h2){3(4h-1)-3h}-2{(1+h8)1/3-(1+3h4)1/2}sinπh=limh→0log(1+h2){3(4h-1)-3h}-2{(1+h24+
)-(1+3h8+ )}sinπh
=limh→0log(1+h2)*{3(4h-1)-3h}-2{-h3+terms containingh2 and
higher powerof h)sinπh
=limh→0log(1+h/2)h/2{3(4h-1h)-3}-{-13+terms containing h and
higher power ofh)}sinπhh
=3loge4-3-(-1/3)π=9π(loge4-1)=9π(loge4-logee)=9πloge(4/e)

hope this helps you

Lim (x approaches1) [(ln(1+x) - ln2)(3.4x-1 - 3x)] / [(7+x)1/3 - (1 + 3x)1/2]sin(x-1)

Lim (x approaches1) [(ln(1+x) - ln2)(3.4^x-1 - 3x)] / [(7+x)^1/3- (1 + 3x)^1/2]sin(x-1)