Question

Let $f\left(x\right)=\frac{\tan x}{x}$, then the value of ${lim}_{x\to 0}{(\left[f\right(x\left)\right]+x^2)}^{\frac{1}{f\left(x\right)}}$ is equal to (where $[.],\{.\}$ denotes greatest integer function and fractional part functions respectively.

• A. $e^{-3}$
• B. $e^3$
• C. $e^2$
• D. $1$

Answer 1

The correct option is D $1$

$\frac{\tan x}{x}=1+\frac{x^2}{31}+....$

$\underset{x\to 0}{lim}\left[f\right(x\left)\right]=1$

$\Rightarrow \underset{x\to 0}{lim}(1+x^2{)}^{\frac{1}{f\left(x\right)}}$

$\Rightarrow e^{\left(\right(1+x^2-1\left)\frac{1}{f\left(x\right)}\right)}\underset{x\to 0}{lim}$

$=e^{\underset{x\to 0}{lim}\frac{x^3}{\tan x}}$

Use L'ospital

$=e^{\underset{x\to 0}{lim}\frac{3x^2}{{\sec }^{2}x}}$

$=e^o$

$=1$.