Question

Let $f\left(x\right)={\left(\frac{\tan x}{x}\right)}^{\frac{1}{x^2}}x\ne 0,f\left(0\right)=0$ .Then

• A. $f$ is continuous at $x=0$
• B. $\underset{x\to 0}{lim}f\left(x\right)=e^{\frac{1}{3}}$
• C. $f$ is discontinuous at $x=0$
• D. $\underset{x\to 0}{lim}f\left(x\right)=0$

Answer 1

The correct option is B $\underset{x\to 0}{lim}f\left(x\right)=e^{\frac{1}{3}}$

$f\left(x\right)={\left(\frac{\tan x}{x}\right)}^{\frac{1}{x^2}}\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}{\left(\frac{\tan x}{x}\right)}^{\frac{1}{x^2}}=\underset{x\to 0}{lim}explog{\left(\frac{\tan x}{x}\right)}^{\frac{1}{x^2}}=exp\underset{x\to 0}{lim}\frac{log\left(\frac{\tan x}{x}\right)}{x^2}$

$=exp\underset{x\to 0}{lim}\frac{x-\tan x+x{\tan }^{2}x}{2x^2\tan x}$ (Applying L-Hospital's rule)

$=exp\underset{x\to 0}{lim}\frac{\tan x(1+{\tan }^{2}x)}{x+2\tan x+x{\tan }^{2}x}$ (Again applying L-Hospital's rule)

$=exp\left(\left(\underset{x\to 0}{lim}(1+{\tan }^{2}x)\right)\left(\underset{x\to 0}{lim}\frac{\tan x}{x+2\tan x+x{\tan }^{2}x}\right)\right)=exp\left(\underset{x\to 0}{lim}\frac{\tan x}{x+2\tan x+x{\tan }^{2}x}\right)$

$=exp\left(\underset{x\to 0}{lim}\frac{{\sec }^{2}x}{1+2{\sec }^{2}x+{\tan }^{2}x+2x\tan x{\sec }^{2}x}\right)$ (Applying L-Hospital's rule)
$=e^{\frac{1}{3}}$