Mathematics - Limits Question with Solution | TestHub

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

${\lim }_{x\to 0}{\left([f(x)]+x^2\right)}^{\frac{1}{\{f(x)\}}}={\lim }_{x\to 0}{\left(1+x^2\right)}^{\frac{1}{\{f(x)\}}}(1^{\infty }$ form $)$

Again, $f(x)=\frac{\tan x}{x}=\frac{x+\frac{x^3}{3}+\frac{2}{15}{x}^5+\dots }{x}$

$$=1+\frac{x^2}{3}+\frac{2}{15}{x}^4+\dots$$

$\{f(x)\}=\frac{x^2}{3}+\frac{2}{15}{x}^4+\dots$

(i) becomes

$${\log }_e\left(e^{\underset{x\to 0}{\lim }{x}^2\times \frac{1}{\left\{f(x)\right\}}}\right)=e^{\underset{x\to 0}{\lim }\frac{x^2}{\frac{x^2}{3}+\frac{2}{15}{x}^4+...}}=3$$

$\therefore$ (C) is the correct answer.