JEE Main 2025 — 7 April, Evening Shift — Question 45

$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\tan (\tan x)-\sin (\sin x)}{(\tan x-\sin x)}$

$=\lim _{x \rightarrow 0} \frac{\tan (\tan x)-\tan x+\sin x-\sin (\sin x)+(\tan x-\sin x)}{\tan x-\sin x}$

$=1+\lim _{x \rightarrow 0}\left(\left(\frac{\tan (\tan x)-\tan x}{\tan ^{3} x}\right) \frac{\tan ^{3} x}{x^{3}}+\left(\frac{\sin x-\sin (\sin x)}{\sin ^{3} x}\right) \frac{\sin ^{3} x}{x^{3}}\right)$ $\cdot \frac{x^{3}}{\tan x-\sin x}$

$=1+\left(\frac{1}{6}+\frac{1}{3}\right) \times \lim _{x \rightarrow 0}\left(\frac{x^{2}}{1-\cos x}\right) \cos x \cdot\left(\frac{x}{\sin x}\right)=1+\frac{1}{2} \cdot 2=2$