Animated Solution for Mathematics - Limits, Continuity and Differentiability: The value of ${\lim }_{x\to 0}\frac{1}{x^3}{\int }_0^x\frac{t\ln (1+t)}{t^4+4}dt$ is

Welcome, future engineer! Today, we are going to unravel a problem that might look intimidating at first glance, but beneath its complex exterior lies a beautiful, elegant structure. We are tasked with finding the value of the limit:

When you first see an integral inside a limit, it is natural to feel a bit of hesitation. But let us pause and look at the big picture. We are dealing with an area function where the integral represents the area under the curve $f(t)=\frac{t\ln (1+t)}{t^4+4}$ from $0$ to $x$.

As $x$ shrinks toward zero, this area collapses into a single point, making the numerator zero. Simultaneously, the denominator $x^3$ also vanishes. We have arrived at the classic $\frac{0}{0}$ indeterminate form, which is our green light to use L'Hopital's Rule.

To apply L'Hopital's Rule, we need to differentiate both the numerator and the denominator. To differentiate the integral, we use the Newton-Leibniz Rule, which states that the derivative of an integral with respect to its upper limit is simply the integrand evaluated at that limit.

The derivative of the numerator is:

Meanwhile, the derivative of our denominator, $x^3$, is a straightforward application of the power rule: $3x^2$. Now, our limit looks much more manageable:

Now, let us simplify the expression. We have an $x$ in the numerator and an $x^2$ in the denominator. Since $x$ is approaching zero but is not zero, we can safely cancel one factor of $x$ to obtain:

Notice how the complexity is melting away. We can separate this into two parts to make the evaluation even clearer:

Do you recognize the first part? It is one of the most fundamental standard limits in calculus: ${\lim }_{x\to 0}\frac{\ln (1+x)}{x}=1$. By isolating this, we have effectively removed the indeterminate nature of the problem.

Now, we are left with the limit of the second part as $x\to 0$:

And there it is! A problem that seemed to involve complex integration and limits has been reduced to a simple fraction. The final answer is $\frac{1}{12}$.