Animated Solution for Mathematics - Differentiation: Evaluate: ${\lim }_{h\to 0}\frac{(a+h{)}^2\sin (a+h)-a^2\sin a}{h}$

Imagine you are standing before a limit that looks like a tangled mess of algebra:

At first glance, it is easy to panic. You might be tempted to expand the sine function, distribute the squares, and get lost in a sea of trigonometric identities.

But stop. Take a breath. In the world of JEE Advanced, complexity is often just a mask for elegance. This limit is not a problem to be brute-forced; it is a definition waiting to be recognized.

Look at the structure again. Does it not scream the First Principle of Derivatives?

Recall that the derivative of a function $f(x)$ at a point $x=a$ is defined as:

This is the geometric heart of calculus—the slope of the tangent line, the limit of the secant line as the gap $h$ vanishes. When you place our given limit side-by-side with this definition, the resemblance is striking.

We have an $h$ in the denominator and a difference of two terms in the numerator. The puzzle pieces are already in place.

Now, let us play detective. We have $f(a)=a^2\sin a$. If we replace the specific point $a$ with a general variable $x$, we uncover our hidden function:

To be absolutely certain, we must verify the first term. If our function is indeed $f(x)=x^2\sin x$, then $f(a+h)$ must be $(a+h{)}^2\sin (a+h)$.

And look—it matches perfectly! We have successfully transformed a terrifying limit into a simple request: find the derivative of $f(x)=x^2\sin x$ at $x=a$.

Now that we know our goal is to find $f^{\prime }(a)$, we turn to our differentiation toolkit. Since $f(x)$ is the product of two functions, $u=x^2$ and $v=\sin x$, we must use the Product Rule:

This is a beautiful dance. We take the first function, $x^2$, and multiply it by the derivative of the second, $\cos x$. Then, we add the second function, $\sin x$, multiplied by the derivative of the first, $2x$.

Putting it all together, we get:

We are at the finish line. We have the general derivative, and now we just need to evaluate it at $x=a$.

Substituting $a$ for $x$, we arrive at our final answer:

By recognizing the derivative definition, we bypassed pages of algebraic expansion and arrived at the truth with elegance and speed. This is the power of conceptual understanding in JEE mathematics—it turns the impossible into the inevitable.