Animated Solution for Mathematics - Definite Integration: ${\lim }_{x\to 0}\frac{{\int }_0^{x^2}\sin \sqrt{t}dt}{x^3}$ is equal to:

Welcome, future engineers! Today, we are going to dissect a problem that often intimidates students at first glance: the limit of an integral. Specifically, we are looking at:

When you see an integral inside a limit, your first instinct might be to panic and try to solve the integral. But stop! Take a deep breath. In JEE Advanced, we are not just testing your ability to integrate; we are testing your ability to see the structure of the problem.

Before we do any heavy lifting, we must check the form of the limit. As $x\to 0$, the upper limit of our integral, $x^2$, also approaches $0$.

This means our integral becomes ${\int }_0^0\sin \sqrt{t}dt$, which is geometrically an area with zero width—it is simply $0$. The denominator, $x^3$, also approaches $0$. We have arrived at the classic $0/0$ indeterminate form, which is our green light to use L'Hopital's Rule.

L'Hopital's Rule states that if we have a $0/0$ form, we can differentiate the numerator and the denominator separately. The denominator is easy: $\frac{d}{dx}(x^3)=3x^2$.

To differentiate the numerator, we use the Newton-Leibniz Rule, which allows us to differentiate an integral with respect to its variable limit:

In our case, $f(t)=\sin \sqrt{t}$, $h(x)=x^2$, and $g(x)=0$. Since $g(x)$ is a constant, its derivative is $0$, which simplifies our life significantly.

Let's apply the rule. We substitute the upper limit $x^2$ into our function $\sin \sqrt{t}$, giving us $\sin \sqrt{x^2}$. We then multiply this by the derivative of the upper limit, which is $\frac{d}{dx}(x^2)=2x$.

So, the derivative of our numerator is $\sin (\sqrt{x^2})\cdot 2x$. Since $\sqrt{x^2}=|x|$, we have $2x\sin |x|$. Now, let's assemble our new limit:

We are almost there! Notice that we have an $x$ in the numerator and an $x^2$ in the denominator. We can cancel one factor of $x$ to obtain:

Since we are taking the limit as $x\to 0$, we can focus on the right-hand limit where $x>0$, allowing us to replace $|x|$ with $x$. The expression becomes:

You should recognize this immediately as the fundamental standard limit: ${\lim }_{x\to 0}\frac{\sin x}{x}=1$. Therefore, our final answer is:

See how the complexity melted away? By trusting the tools—L'Hopital's Rule and the Newton-Leibniz Rule—we turned a terrifying integral into a simple, elegant result. Keep practicing this structural thinking, and you will dominate the JEE Advanced!