Question

Evaluate the following limits:
$\underset{x\to {\frac{\pi }{2}}^{-}}{lim}(tanx{)}^{cosx}$

Answer 1

$L={lim}_{x\to {\frac{\pi }{2}}^{-}}(\tan x{)}^{\cos x}$

Taking log at both sides

$\ln L={lim}_{x\to {\frac{\pi }{2}}^{-}}\cos x\ln \tan x$

As $x\to \frac{\pi }{2},\tan x\to +\infty$ So,$\ln (\tan x)\to \infty$. On the other hand $\cos x$ coverges to $0$.

${lim}_{x\to {\frac{\pi }{2}}^{-}}\cos x\ln \left(\tan x\right)={lim}_{x\to {\frac{\pi }{2}}^{-}}\frac{\ln \left(\tan x\right)}{\frac{1}{\cos x}}$

Using L'Hospitals rule,

$\Rightarrow {lim}_{x\to {\frac{\pi }{2}}^{-}}\frac{\frac{1}{\tan \left(x\right)}.\frac{1}{{\cos }^2\left(x\right)}}{-\frac{1}{{\cos }^2\left(x\right)}.(-\sin (x\left)\right)}$

$\Rightarrow {lim}_{x\to {\frac{\pi }{2}}^{-}}\frac{\cos \left(x\right)}{{\sin }^2\left(x\right)}=0$

Thus, $\ln \left(L\right)=0$

$\therefore L=e^0$

$L=1$

Thus, ${lim}_{x\to {\frac{\pi }{2}}^{-}}(\tan x{)}^{\cos x}=1$