Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=\{\begin{array}{ll}(x-1{)}^2\sin \frac{1}{(x-1)}-|x|& \text{if }x\ne 1\\ -1& \text{if }x=1\end{array}$ be a real-valued function. Then the set of points where $f(x)$ is not differentiable is .........

Unmasking Non-Differentiability

Welcome, fellow explorer of the mathematical universe! Today, we are going to act as detectives. We have a function, $f(x)$, that looks like a chaotic mess of oscillations and sharp corners.

Our mission? To find the exact points where this function refuses to be smooth—where it is not differentiable.

Identifying the Suspects

Let us look at our function:

In the world of calculus, we do not need to check every point on the real number line. We only hunt for the troublemakers.

Looking at this expression, two suspects immediately jump out. First, $x=1$, because the function definition changes there and we have that wild $\sin \frac{1}{x-1}$ term. Second, $x=0$, because of the absolute value term $|x|$, which we know is the classic 'sharp corner' culprit.

The Mystery of $x=1$

Before we can even discuss the derivative at $x=1$, we must ensure the function is continuous. If the graph has a jump or a hole, it cannot be differentiable.

Let us calculate the limit as $x$ approaches $1$:

As $x$ gets closer to $1$, the term $\sin \frac{1}{x-1}$ oscillates violently between $-1$ and $1$. It is a chaotic mess!

But look at the term multiplying it: $(x-1{)}^2$. As $x\to 1$, this term shrinks to zero. By the Squeeze Theorem, zero times any bounded oscillating value is simply zero.

The chaos is damped out! So, the limit becomes $0-|1|=-1$. Since the function value $f(1)$ is given as $-1$, the limit equals the function value. The bridge is intact; the function is continuous at $x=1$.

Now, is it smooth? We use the first principle of derivatives:

Substituting $x=1+h$, we get:

Since $h$ is tiny, $1+h$ is positive, so $|1+h|=1+h$. The expression simplifies beautifully:

Again, the Squeeze Theorem tells us $h\sin \frac{1}{h}\to 0$. We are left with a finite slope of $-1$. The function is perfectly differentiable at $x=1$! Our first suspect is innocent.

The Sharp Corner at $x=0$

Now, let us turn our attention to $x=0$. We can split our function into two parts: $g(x)=(x-1{)}^2\sin \frac{1}{x-1}$ and $h(x)=|x|$.

Around $x=0$, $g(x)$ is a smooth, well-behaved function. It is differentiable.

However, $h(x)=|x|$ is the classic example of a function that is continuous but not differentiable at the origin. It has a sharp 'V' shape.

There is a powerful theorem in calculus: if you add a differentiable function to a non-differentiable one, the result is non-differentiable. The sharp corner 'infects' the entire sum. Therefore, $f(x)$ is not differentiable at $x=0$.

We have cleared $x=1$ and convicted $x=0$. The set of points where $f(x)$ is not differentiable is simply $\{0\}$.

Remember this lesson: always look for the damping factors that tame oscillations, and never underestimate the power of a sharp corner to break differentiability. Keep practicing, and you will master these mysteries in no time!