Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $g(x)=xf(x)$, where $f(x)=\{\begin{array}{ll}x\sin \frac{1}{x},& x\ne 0\\ 0,& x=0\end{array}$. At $x=0$

A JEE Masterclass

Welcome, future engineers! Today, we are diving into one of the most beautiful and subtle corners of calculus. We are going to dissect the behavior of two functions, $f(x)=x\sin \frac{1}{x}$ and $g(x)=x^2\sin \frac{1}{x}$, at the critical point $x=0$.

This problem is a rite of passage for every JEE Advanced aspirant. It tests your ability to look past the algebraic surface and see the geometric soul of a function.

The Failure of $f(x)$

Imagine you are standing at the origin, $x=0$. You want to know if the function $f(x)=x\sin \frac{1}{x}$ has a slope there.

To find out, we must use the First Principle of Derivatives. We define the derivative at zero as:

Substituting our function, we get:

Now, pause and visualize this. As $h$ gets smaller and smaller, $\frac{1}{h}$ rushes toward infinity. The sine function, trapped between $-1$ and $1$, will oscillate infinitely fast as it approaches the origin.

Because it never settles on a single value, the limit does not exist. Thus, $f(x)$ is not differentiable at $x=0$. It is a jagged, broken path at the origin.

The Rescue by $g(x)$

Now, let us look at $g(x)=x^2\sin \frac{1}{x}$. Does the extra power of $x$ change the story?

Let us apply the same First Principle:

Here is where the magic happens. We know that $-1\le \sin \frac{1}{h}\le 1$. If we multiply this entire inequality by $h$ (assuming $h>0$), we get:

As $h$ approaches zero, both $-h$ and $h$ squeeze toward zero. By the Squeeze Theorem, our limit is forced to be $0$.

This means $g(x)$ is differentiable at $x=0$, and its slope is exactly $0$. The extra factor of $x$ has tamed the wild oscillation of the sine function!

The Subtle Trap of Continuity

We have established that $g(x)$ is differentiable at $x=0$. But is its derivative, $g^{\prime }(x)$, continuous? This is where many students stumble.

To check continuity, we need to see if ${\lim }_{x\to 0}{g}^{\prime }(x)=g^{\prime }(0)$. First, we find $g^{\prime }(x)$ for $x eq 0$ using the product rule:

This simplifies to:

Now, take the limit as $x\to 0$. The first term, $2x\sin \frac{1}{x}$, goes to $0$ (again, thanks to the Squeeze Theorem).

However, the second term, $-\cos \frac{1}{x}$, oscillates between $-1$ and $1$ as $x\to 0$. It never settles! Therefore, the limit of $g^{\prime }(x)$ does not exist.

Since the limit does not exist, $g^{\prime }(x)$ is not continuous at $x=0$.

What a journey! We have seen that $f(x)$ is not differentiable, $g(x)$ is differentiable, but its derivative is not continuous.

This problem teaches us that differentiability is a local property, while continuity of the derivative is a global property of the function's slope. Keep visualizing these limits, keep trusting the Squeeze Theorem, and you will master the calculus of the JEE Advanced!