# Evaluate the limit for the following if exists. $\;\lim\limits_{x \to 0}(\cos x)^{\large\frac{1}{x}}$

Toolbox:
• L'Hopital's rule: Let $f$ and $g$ be continous real valued functions defined on the closed interval $[a,b], f,g$ be differentiable on $(a,b)$ and $g'(c) \neq 0$
• Then if $\lim\limits_{x \to c}\; f(x)=0, \lim \limits_{x \to c}\; g(x)=0$ and
• $\lim\limits_{x \to c} \large\frac{f'(x)}{g'(x)}$$=L, it follows that • \lim \limits_{x \to c} \large\frac{f(x)}{g(x)}$$=L$
• The conditions for L'Hopital's rule remains unchanged if $\lim\limits_{x \to c} f(x)=\pm \infty$ and $\lim\limits_{x \to c} g(x)=\pm \infty$
• All indeterminate forms can be reduced t $\large\frac{0}{0}$ or $\large\frac{\infty}{\infty}$
Step 1:
$\;\lim\limits_{x \to 0}(\cos x)^{\large\frac{1}{x}}$ this is of the form $1^{\infty}$
Step 2:
$\log _eL=\log _e \lim\limits_{x \to 0}\cos x^{\large\frac{1}{x}};$ (By composite function theroem)
$\log _eL=\lim\limits_{x \to 0}\log _e \cos x^{\large\frac{1}{x}}$
Step 3:
$\log _e L= \lim \limits_{x \to 0} \large\frac{1}{x} $$\log _e \cos x which is of the form \large\frac {-\infty}{\infty} Step 4: Applying L'Hopital's rule, we have \log _e L= \lim \limits_{x \to 0} \large\frac{\Large\frac{1}{\cos x} (-\sin x)}{1} \qquad= \lim \limits_{x \to 0} -\tan x =0 Step 5: \therefore L=1=> \lim \limits_{x \to 0} \cos x^{\large\frac{1}{x}}$$=1$

edited Jul 29, 2013 by meena.p