Question

$\displaystyle L=\lim_{x\to0}(\sin x)^{\tan x}$ is equal to? ​

Answer 1

refer the above attachment

hope it helpful..

Answer 2

$\: \: \: \: \:{\large{\pmb{\sf{\underline{ Here's \: your \: required \: solution!! }}}}}$

$\sf{ \displaystyle:\implies \red { L=\lim_{x\to0}(\sin x)^{\tan x}}}$

Taking log both sides :-

$\sf {\displaystyle:\implies\log L=\lim_{x\to0}\tan x\cdot\log(\sin x)}$

$\sf {\displaystyle:\implies\log L=\lim_{x\to0}\dfrac{\log(\sin x)}{\cot x}}$

Applying L'hospital's Rule:-

$\sf {\displaystyle:\implies\log L=\lim_{x\to0}\dfrac{\left(\dfrac{\cos x}{\sin x}\right)}{-\csc^2x}}$

$\sf {\displaystyle:\implies\log L=-\lim_{x\to0}\dfrac{\left(\dfrac{\sin x\cos x}{\sin^2x}\right)}{\left(\dfrac{1}{\sin^2x}\right)}}$

$\sf {\displaystyle:\implies\log L=-\lim_{x\to0}\sin x\cos x}$

$\sf \displaystyle:\implies\log L=-\dfrac{1}{2}\lim_{x\to0}\sin(2x)$

$\sf {\displaystyle:\implies\log L=-\dfrac{1}{2}\sin(2\times 0)}$

$\sf {\displaystyle:\implies\log L=0}$

$\sf {\displaystyle:\implies L=e^0}$

$\sf{ \displaystyle:\implies\red {\underline{L=1}}}$