Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to 0}\frac{{\int }_0^{x}t\sin (10t)dt}{x}$ is equal to:

Welcome, future engineers! Today, we are going to dissect a problem that might look intimidating at first glance, but reveals a beautiful simplicity once we apply the right tools. We are tasked with evaluating the limit:

At first, your instinct might be to dive into the integral, performing integration by parts on $t\sin (10t)$. But pause for a moment. In the world of JEE Advanced, we don't just solve; we strategize.

Let us look at the structure. As $x$ approaches $0$, the upper limit of our integral, $x$, meets the lower limit, $0$.

The area under the curve $f(t)=t\sin (10t)$ from $0$ to $0$ is, by definition, $0$. Simultaneously, the denominator $x$ also approaches $0$.

We have arrived at the classic $\frac{0}{0}$ indeterminate form. This is not a dead end; it is an invitation to use L'Hospital's Rule.

To apply L'Hospital's Rule, we need the derivative of the numerator. This is where many students stumble.

We are differentiating an integral with respect to its upper limit. We invoke the Leibniz Rule, which states:

Applying this to our numerator, the derivative of ${\int }_0^{x}t\sin (10t)dt$ is simply $x\sin (10x)$. The lower limit, being a constant, contributes nothing to the derivative.

Meanwhile, the derivative of our denominator, $x$, is simply $1$.

Now, our limit has transformed into something much friendlier:

We no longer have an indeterminate form. We can proceed with direct substitution.

Substituting $x=0$ into the expression, we get $0\cdot \sin (10\cdot 0)$, which is $0\cdot \sin (0)$. Since $\sin (0)=0$, the entire expression collapses to 0.

We have navigated the complexity and arrived at the solution with precision. Remember, the key to mastering calculus is not just knowing the formulas, but understanding when to deploy them to simplify the path ahead.