IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

salonee_hs

If lim xtendsto 0[cot(pie/4+x)]^1/x= Ae²then A=?

e²

e^-4

e

e³

ans is b

studyid

$\mathop{\lim}\limits_{{0} \to {0}}[{cot(\frac{\pi}{4}+x)}]^{\frac{1}{x}}$

$= \mathop{\lim}\limits_{{x} \to {0}}[{\frac{1}{tan(\frac{\pi}{4}+x)}}]^{\frac{1}{x}}$

$= \mathop{\lim}\limits_{{x} \to {0}}[{\frac{1-tanx}{1+tanx}}]^{\frac{1}{x}}$

$= \mathop{\lim}\limits_{{x} \to {0}}[{1+(\frac{-2tanx}{1+tanx})}]^{\frac{1+tanx}{-2tanx}(\frac{-2tanx}{x})(\frac{1}{1+tanx})}$

Applying the Limits we have this limit as ${e}^{-2}$

and ${e}^{-2} = A {e}^{2}$

Hence $A = {e}^{-4}$

rohitkuruvila

ans is b.

see,[cot(pie/4+x)]^1/xcan be written as (cot x - 1)/(1 + cot x) (expand cot(A+B)=(cotAcotB - 1)/cotB+cotA)

[ (cot x - 1)/(1 + cot x) ]^1/x

{ 1 + [-2/(1+cot x)] }^1/x

use (1 + f(x))^1/x=e

{ 1 + [-2/(1+cot x)] }^{[(1+cot x)/-2]} ^{-2/x(cot x + 1)}

e^-2=Ae²

A=e^-4