Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=\frac{1-|x|(1+|1-x|)}{|1-x|}\cos \left(\frac{1}{1-x}\right)$ for $x\ne 1$. Then

Welcome, future engineer. Today, we stand before a function that seems designed to intimidate. Look at it:

It has absolute value signs, a rational structure, and a trigonometric term that screams 'singularity' at $x=1$. In the JEE Advanced arena, this is a classic test of your composure.

When you see a function like this, do not panic. The secret is not to solve it all at once, but to dismantle it piece by piece.

The Modulus Trap (LHL)

Our first task is to understand the behavior of the modulus $|1-x|$ as we approach $x=1$ from the left ($x\to 1^{-}$). When $x$ is slightly less than $1$, the quantity $1-x$ is positive. Therefore, the modulus $|1-x|$ simply opens as $1-x$.

Similarly, since $x$ is positive, $|x|=x$. Substituting these into our function, we get:

Now, let's simplify the inner bracket: $1+(1-x)=2-x$. Our numerator becomes $1-x(2-x)=1-2x+x^2$.

Stop and look at that expression. It is a perfect square! It is $(1-x{)}^2$. So, our function simplifies to:

Canceling the common factor $(1-x)$, we are left with $f(x)=(1-x)\cos \left(\frac{1}{1-x}\right)$.

The Squeeze Theorem

Now, evaluate the limit as $x\to 1^{-}$. The term $(1-x)$ approaches $0$. The term $\cos \left(\frac{1}{1-x}\right)$ oscillates wildly between $-1$ and $1$.

Here, we invoke the Squeeze Theorem: $0$ multiplied by any bounded value is $0$. Thus, the Left-Hand Limit is $0$. We have conquered the first half.

The RHL Shift

Now, we approach from the right ($x\to 1^{+}$). Here, $x>1$, so $1-x$ is negative. The modulus $|1-x|$ must open as $-(1-x)=x-1$.

Our function becomes:

The inner bracket simplifies to $1+x-1=x$. The numerator is now $1-x(x)=1-x^2$. We factorize this as $-(x-1)(x+1)$.

The Divergence

Substituting this back, we have:

The $(x-1)$ terms cancel, leaving us with $-(x+1)\cos \left(\frac{1}{1-x}\right)$. As $x\to 1^{+}$, the term $-(x+1)$ approaches $-2$.

But the cosine term still oscillates between $-1$ and $1$. The product oscillates between $-2$ and $2$.

Because it never settles on a single value, the limit does not exist. You have successfully navigated the trap. Keep this rigor, and no function will ever defeat you.