Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=\{\begin{array}{ll}{x}^2\sin \left(\frac{1}{x}\right)& ,x\ne 0\\ 0& ,x=0\end{array}$; Then at $x=0$

Welcome, students. Today, we are diving into a function that is a rite of passage for every serious JEE aspirant. We are looking at $f(x)=x^2\sin (\frac{1}{x})$ for $x eq 0$, and $f(0)=0$.

This is not just a function; it is a lesson in humility. It teaches us that our intuition about continuity and differentiability can be deceptive. Let us peel back the layers of this problem together.

First, we must establish continuity at $x=0$. You might be tempted to just plug in $x=0$, but the term $\sin (\frac{1}{x})$ is undefined there.

As $x$ gets closer to zero, $\frac{1}{x}$ explodes to infinity, and the sine function oscillates wildly between $-1$ and $1$. How do we tame this chaos? We use the Squeeze Theorem.

We know that $-1\le \sin (\frac{1}{x})\le 1$. If we multiply this entire inequality by $x^2$, we get:

As $x\to 0$, both $-x^2$ and $x^2$ collapse to zero. Thus, our function is squeezed into the origin. Since the limit is $0$ and $f(0)=0$, the function is continuous. We have survived the first hurdle.

Now, is it differentiable? Many students try to differentiate $x^2\sin (\frac{1}{x})$ using the product rule and then plug in $x=0$. Do not do this!

The product rule assumes the derivative exists in a neighborhood. At $x=0$, we must use the definition:

Substituting our function, we get:

Again, the Squeeze Theorem comes to our rescue. Since $|h\sin (\frac{1}{h})|\le |h|$, as $h\to 0$, the limit is $0$. So, $f^{\prime }(0)=0$. The function is differentiable at the origin!

Now, let us find the derivative for $x eq 0$. Here, we are on safe ground to use the Product Rule: $\frac{d}{dx}[u\cdot v]=u^{\prime }v+u{v}^{\prime }$.

Let $u=x^2$ and $v=\sin (\frac{1}{x})$. The derivative is:

Simplifying this, the $x^2$ terms cancel, leaving us with:

Finally, we ask: is $f^{\prime }(x)$ continuous at $x=0$? We need to check if ${\lim }_{x\to 0}{f}^{\prime }(x)=f^{\prime }(0)$.

We know $f^{\prime }(0)=0$. Let us evaluate:

The first term, $2x\sin (\frac{1}{x})$, goes to $0$ by the Squeeze Theorem. But look at the second term: $-\cos (\frac{1}{x})$.

As $x\to 0$, $\frac{1}{x}\to \infty$, and $\cos (\frac{1}{x})$ oscillates forever. It never approaches a single value. Therefore, the limit of the derivative does not exist.

We have proven that while the function is differentiable, its derivative is not continuous. This is the beauty of calculus—a function can be smooth enough to have a tangent, but its slope can be too erratic to be continuous. Keep this example in your toolkit; it is a classic JEE trap!