JEE Main 2026 — Limits Question with Solution

The given function is $f(x)={\lim }_{y\to 0}\frac{(1-\cos (xy))\tan (xy)}{y^3}$.

Multiplying and dividing by $x^3$, we get:

$f(x)={\lim }_{y\to 0}\frac{1-\cos (xy)}{(xy{)}^2}\cdot \frac{\tan (xy)}{xy}\cdot x^3$

Using the standard limits ${\lim }_{t\to 0}\frac{1-\cos t}{t^2}=\frac{1}{2}$ and ${\lim }_{t\to 0}\frac{\tan t}{t}=1$, we obtain:

$f(x)=\frac{1}{2}\cdot 1\cdot x^3=\frac{x^3}{2}$

We need to find the number of solutions for the equation $f(x)=\sin x$, which is $\frac{x^3}{2}=\sin x$.

Let $g(x)=\frac{x^3}{2}-\sin x$.

Differentiating with respect to $x$:

$g^{\prime }(x)=\frac{3x^2}{2}-\cos x$

$g^{\prime \prime }(x)=3x+\sin x$

For $x>0$, $g^{\prime \prime }(x)>0$, which implies that $g^{\prime }(x)$ is strictly increasing on $(0,\infty )$.

We have $g^{\prime }(0)=-1<0$ and $g^{\prime }(\pi /2)=\frac{3{\pi }^2}{8}>0$. Thus, $g^{\prime }(x)=0$ has exactly one root in $(0,\infty )$, say at $x=x_0$.

This means $g(x)$ decreases on $(0,x_0)$ and increases on $(x_0,\infty )$.

Since $g(0)=0$ and $g(x)$ initially decreases, $g(x_0)<0$. As $x\to \infty$, $g(x)\to \infty$. Therefore, $g(x)=0$ has exactly one positive root.

Since $g(x)$ is an odd function ($g(-x)=-g(x)$), it must also have exactly one negative root.

Including the trivial root $x=0$, the equation $g(x)=0$ has exactly $3$ solutions.

Answer: $3$