Question

If $L=\underset{x\to 0}{lim}{\left(\frac{\tan x}{x}\right)}^{1/x^2}$, then the value of $\frac{1}{\ln L}$ is

Answer 1

$L=\underset{x\to 0}{lim}{\left(\frac{\tan x}{x}\right)}^{1/x^2}\left[1^{\infty }\text{form}\right]$
Now, taking log on both sides, we get
$\Rightarrow \ln L=\underset{x\to 0}{lim}\frac{\ln \left(\frac{\tan x}{x}\right)}{x^2}\Rightarrow \ln L=\underset{x\to 0}{lim}\frac{\ln \tan x-\ln x}{x^2}$
Using L'Hospital's Rule,

Differntiate with respect to '$x$' we get

$\Rightarrow \ln L=\underset{x\to 0}{lim}\frac{\frac{{\sec }^{2}x}{\tan x}-\frac{1}{x}}{2x}\Rightarrow \ln L=\underset{x\to 0}{lim}\frac{x-\sin x\cos x}{2x^2\sin x\cos x}\Rightarrow \ln L=\underset{x\to 0}{lim}\frac{2x-\sin 2x}{4x^3}$
Using L'Hospital's Rule,

Differntiate with respect to '$x$' we get
$\Rightarrow \ln L=\underset{x\to 0}{lim}\frac{2-2\cos 2x}{12x^2}\Rightarrow \ln L=\underset{x\to 0}{lim}\frac{2{\sin }^{2}x}{6x^2}\Rightarrow \ln =\frac{2}{6}\underset{x\to 0}{lim}\frac{{\sin }^{2}x}{x^2}\ln L=\frac{1}{3}\left(1\right)\therefore \frac{1}{\ln L}=3$