JEE Main 2026 — 5 April, Evening Shift — Question 41

Given $f(x) = \lim_{y \to 0} \frac{(1-\cos(xy)) \tan(xy)}{y^3}$. Use standard limits: $\lim_{u \to 0} \frac{1-\cos u}{u^2} = \frac12$ and $\lim_{u \to 0} \frac{\tan u}{u} = 1$. Rewrite: $f(x) = \lim_{y \to 0} \frac{1-\cos(xy)}{(xy)^2} \cdot \frac{\tan(xy)}{xy} \cdot \frac{(xy)^3}{y^3}$. As $y \to 0$, $xy \to 0$, so $f(x) = \frac12 \cdot 1 \cdot x^3 = \frac{x^3}{2}$. Solve $\frac{x^3}{2} = \sin x$. Consider $g(x) = \frac{x^3}{2} - \sin x$. $g(0) = 0$, so $x=0$ is a solution. For $x>0$, $g(\pi/2) = (\pi/2)^3/2 - 1 > 0$, $g(\pi) = \pi^3/2 - 0 > 0$, but $g(2) = 4 - \sin 2 > 0$. Check $g(1) = 0.5 - \sin 1 \approx 0.5 - 0.84 < 0$.

So there is a root in $(0,1)$. For $x<0$, by symmetry, there is a root in $(-1,0)$. Thus three solutions: $x=0$,

one positive, one negative.