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Evaluate the limit for the following if exists. $\;\lim\limits_{x \to \large\frac{\pi}{2}\;-} (\tan x)^{\cos x}$

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  • 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 $\large\frac{\infty}{\infty}$
$\;\lim\limits_{x \to \Large\frac{\pi}{2}\;-} (\tan x)^{\cos x}$ is of the form $ \infty$
Step 2:
Let $L=\lim\limits_{x \to \Large\frac{\pi}{2}\;-} (\tan x)^{\cos x}$
$\therefore$ $\log _cL=\log _e \lim \limits_{x \to \Large\frac{\pi}{2} \; -} \tan x ^{\cos x}$ (By composite function theroem)
Step 3:
$\log _cL=\log _e \lim \limits_{x \to \Large\frac{\pi}{2} \; -} \cos x \log _c \tan x$ which is of the form $0 \times \infty$
This can be written as
$\log _eL=\log _e \lim \limits_{x \to \Large\frac{\pi}{2} \; -} \large\frac { \log _c \tan x}{\sec x }$
which is of the form $\large\frac {\infty}{\infty}$
Step 4:
Applying L'Hopital's rule, we have
$\log _eL=\lim\limits_{x \to \Large\frac{\pi}{2} \; -} \large\frac { \Large\frac{1}{\tan x} \large \sec ^2 x}{\sec x \tan x}$$=\lim\limits_{x \to \frac{\pi}{2}\;-} \large\frac{\sec x }{\tan ^2x}$
Step 5:
Applying L'Hopital's rule once again
$\log _eL=\lim\limits_{x \to \Large\frac{\pi}{2} \; -} \large\frac{\sec x \tan x}{2 \tan x .\sec ^2 x}$
$\qquad =\lim\limits_{x \to \Large\frac{\pi}{2} \; -} \large \frac{1}{2 \sec x}$
$\qquad =\lim\limits_{x \to \Large\frac{\pi}{2} \; -} \large \frac{1}{2}$$ \cos x=0$
Step 6:
$\therefore \log _e L=0 =>L=1$
$\;\lim\limits_{x \to \Large\frac{\pi}{2}\;-} (\tan x)^{\cos x}$$=1$


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

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