JEE Main 2025 Apr 7 Shift 2, Mathematics Q21: Methods to Evaluate Limits

Split the numerator by adding and subtracting $\tan x$ and $\sin x$: $\tan (\tan x)-\sin (\sin x)=[\tan (\tan x)-\tan x]+[\tan x-\sin x]+[\sin x-\sin (\sin x)]$.

Dividing throughout by $x^3$ and taking the limit as $x\to 0$, three standard small-angle limits apply: $\lim \frac{\tan u-u}{u^3}=\frac{1}{3}$ (with $u=\tan x\to 0$, and ${\tan }^{3}x/x^3\to 1$), $\lim \frac{\tan x-\sin x}{x^3}=\frac{1}{2}$, and $\lim \frac{u-\sin u}{u^3}=\frac{1}{6}$ (with $u=\sin x\to 0$).

So the numerator, divided by $x^3$, tends to $\frac{1}{3}+\frac{1}{2}+\frac{1}{6}=1$.

The denominator, divided by $x^3$, tends to $\frac{1}{2}$ (a standard limit).

$\therefore f(0)=\underset{x\to 0}{\lim }\frac{\text{numerator}/x^3}{\text{denominator}/x^3}=\frac{1}{1/2}=2$.