Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=\{\begin{array}{ll}(x-1)\sin \frac{1}{x-1},& \text{if }x\ne 1\\ 0,& \text{if }x=1\end{array}$. Then which one of the following is true?

Welcome, future IITians! Today, we are going to dissect a classic JEE problem that tests your fundamental understanding of calculus. We are looking at a function that seems simple at first glance but hides a fascinating, 'pathological' behavior at a specific point.

Let's dive into the world of the function defined as:

The Envelope of Oscillation

Imagine you are standing on the coordinate plane. For any $x eq 1$, our function is the product of a linear term $(x-1)$ and an oscillating term $\sin \frac{1}{x-1}$.

The linear term $(x-1)$ acts as an 'envelope.' As $x$ approaches $1$, the value of $(x-1)$ shrinks toward zero. This forces the entire function to be 'squeezed' between the lines $y=(x-1)$ and $y=-(x-1)$.

Even though the sine function is oscillating wildly, the $(x-1)$ term dampens these oscillations, forcing the function to pass through the origin $(1,0)$. This is why the function is continuous at $x=1$. But does continuity imply differentiability? Let's find out.

The Battle at $x=1$

To check if $f(x)$ is differentiable at $x=1$, we cannot simply differentiate the expression. We must return to the bedrock of calculus: the definition of the derivative.

The derivative $f^{\prime }(1)$ exists if and only if the limit ${\lim }_{h\to 0}\frac{f(1+h)-f(1)}{h}$ exists and is finite. Let's substitute our function into this limit. Since $f(1)=0$, the expression becomes:

Simplifying the numerator, we get $h\sin \frac{1}{h}$. Now, look at the beauty of the algebra: the $h$ in the numerator and the $h$ in the denominator cancel out perfectly!

Here is where the trap lies. As $h\to 0$, the argument $\frac{1}{h}$ approaches infinity. The sine of infinity does not settle down to a single value; it oscillates infinitely fast between $-1$ and $1$.

Because the limit does not approach a unique, finite value, the derivative at $x=1$ simply does not exist. The function is continuous, but it is not differentiable.

Smooth Sailing at $x=0$

Now, let's contrast this with the point $x=0$. Unlike $x=1$, the point $x=0$ is far from the singularity. The function is perfectly well-behaved here.

We can use the standard rules of differentiation, specifically the Product Rule. Let $f(x)=(x-1)\sin \frac{1}{x-1}$. Applying the product rule, we get:

Using the chain rule on the second term, the derivative of $\sin \frac{1}{x-1}$ is $\cos \frac{1}{x-1}\cdot \left(-\frac{1}{(x-1{)}^2}\right)$. Substituting this back, we get:

Simplifying the second term, the $(x-1)$ in the numerator cancels with one factor of $(x-1)$ in the denominator, leaving us with:

This formula is valid for all $x eq 1$. Since $x=0$ is not $1$, we can safely plug it in:

Since $\cos (1)-\sin (1)$ is a unique, finite real number, the derivative exists at $x=0$.

We have successfully navigated the treacherous waters of this piecewise function. We found that while the function is continuous everywhere, the infinite oscillation at $x=1$ destroys the possibility of a unique tangent line, making it non-differentiable at that point.

However, at $x=0$, the function is perfectly smooth. This confirms that $f$ is differentiable at $x=0$ but not at $x=1$. Keep this distinction in mind—differentiability is a local property, and it demands much more than just continuity!