Animated Solution for Mathematics - Limits, Continuity and Differentiability: The set of all points where the function $f(x)=\frac{x}{1+|x|}$ is differentiable, is

A Lesson in Skepticism

Welcome, future engineer. Today, we are going to dismantle one of the most notorious traps in the JEE Advanced syllabus: the 'Modulus Phobia.'

When you see a function like $f(x)=\frac{x}{1+|x|}$, your brain might immediately scream, 'Sharp corner! Non-differentiable at $x=0$!' It is a natural reaction.

We are trained to recognize the V-shape of $|x|$ and associate it with a point where the derivative fails to exist. But in the world of JEE, intuition is a tool, not a rule. Let us look past the surface and dive into the mechanics of this function.

Deconstructing the Beast

To understand this function, we must strip away the modulus. The modulus function $|x|$ is a chameleon; it changes its identity based on the sign of $x$.

We must define our function in two distinct regions:

1. When $x\ge 0$, $|x|=x$. Our function becomes $f(x)=\frac{x}{1+x}$.

2. When $x<0$, $|x|=-x$. Our function becomes $f(x)=\frac{x}{1-(-x)}=\frac{x}{1-x}$.

Now, we have two clean, rational functions. They are smooth, continuous, and perfectly well-behaved in their respective domains. The only place where the 'seam' exists—the only place where these two definitions meet—is at $x=0$. This is our battlefield.

The Continuity Check

Before we even talk about derivatives, we must ensure the function is continuous. A function cannot be differentiable if it is broken.

Let us check the limit at $x=0$ from both sides:

- $f(0)=\frac{0}{1+0}=0$

- ${\lim }_{x\to 0^{+}}f(x)={\lim }_{x\to 0^{+}}\frac{x}{1+x}=0$

- ${\lim }_{x\to 0^{-}}f(x)={\lim }_{x\to 0^{-}}\frac{x}{1-x}=0$

Since the Left Hand Limit (LHL), the Right Hand Limit (RHL), and the value of the function at $x=0$ are all equal to $0$, the function is continuous. The bridge is intact. Now, we can proceed to the derivative.

The Differentiability Showdown

This is where the magic happens. We need to calculate the derivative at $x=0$ using the first principles definition. We need to see if the slope from the right matches the slope from the left.

The Right Hand Derivative (RHD): We use the definition $f^{\prime }(0^{+})={\lim }_{h\to 0^{+}}\frac{f(0+h)-f(0)}{h}$. Substituting our function for $x>0$, we get:

The Left Hand Derivative (LHD): We use the definition $f^{\prime }(0^{-})={\lim }_{h\to 0^{+}}\frac{f(0-h)-f(0)}{-h}$. Substituting our function for $x<0$, we get:

Look at that! The RHD is $1$, and the LHD is $1$. They are identical. The slope from the left perfectly matches the slope from the right.

There is no sharp corner. There is no kink. The function transitions through the origin with a smooth, continuous slope of $1$.

For all other points where $x eq 0$, the function is a simple rational function, which is differentiable everywhere in its domain. Since the denominator $1+|x|$ is always $\ge 1$, it never vanishes, meaning there are no asymptotes to worry about.

So, what is the set of all points where $f(x)$ is differentiable? It is the entire real line, $(-\infty ,\infty )$.

This problem is a beautiful reminder of why we study calculus. It teaches us to be rigorous. It teaches us that while patterns exist, they are not universal laws.

Never let a symbol like $|x|$ dictate your answer before you have done the math. Trust the process, trust the limits, and you will never fall into the trap. Keep practicing, keep questioning, and keep pushing forward!