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

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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$

 

answered Jul 29, 2013 by meena.p
edited Jul 29, 2013 by meena.p
 

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