Animated Solution for Mathematics - Limits, Continuity and Differentiability: The value of ${\lim }_{x\to 0}\frac{{\int }_0^{x^2}{\sec }^{2}tdt}{x\sin x}$ is

We are tasked with evaluating the following limit:

As $x\to 0$, the upper limit of the integral $x^2$ approaches $0$. Since the integral of a finite function from $0$ to $0$ is $0$, and the denominator $x\sin x$ also approaches $0$, we have identified the classic $\frac{0}{0}$ indeterminate form.

To resolve this, we apply L'Hôpital's Rule, which requires differentiating the numerator and the denominator. For the numerator, we utilize the Leibniz Integral Rule:

Here, $f(t)={\sec }^{2}t$ and $g(x)=x^2$. Applying this rule, the derivative of the numerator becomes:

Next, we differentiate the denominator $x\sin x$ using the Product Rule:

Substituting these derivatives back into our limit, we obtain:

Substituting $x=0$ at this stage still yields $\frac{0}{0}$. Instead of differentiating again, we simplify the expression by dividing both the numerator and the denominator by $x$:

We now apply the fundamental trigonometric limit ${\lim }_{x\to 0}\frac{\sin x}{x}=1$. As $x\to 0$, we observe that ${\sec }^2(x^2)\to {\sec }^2(0)=1$ and $\cos x\to \cos (0)=1$.

Substituting these values into our simplified expression, we get:

The complexity of the original expression has dissolved, revealing the final result. The value of the limit is $1$.