JEE Main 2026 Apr 5 Shift 2, Mathematics Q17: Methods to Evaluate Limits

Use standard limits: $1-\cos (xy)\sim \frac{(xy{)}^2}{2}$ and $\tan (xy)\sim xy$ as $y\to 0$.

So $f(x)={\lim }_{y\to 0}\frac{(xy{)}^2/2\cdot xy}{y^3}=\frac{x^3}{2}$.

The equation becomes $\frac{x^3}{2}=\sin x$. The cubic $x^3/2$ and $\sin x$ both vanish at $x=0$ and are odd; they intersect at $x=0$ and at one pair of symmetric points (near where $|x^3/2|=|\sin x|$ before the cubic outruns the sine).

$\therefore$ Number of solutions $=3$.