Q.Find all points of discontinuity of $f$, where $f$ is defined by $f(x) = \begin{cases} \frac{|x|}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$

Concept: Continuity At A Point — A function is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.

Step 1: For $x \neq 0$, $\frac{|x|}{x}$ equals $1$ when $x > 0$ and $-1$ when $x < 0$.

Step 2: The left-hand limit at $0$ is $\lim_{x \to 0^-} f(x) = -1$. The right-hand limit is $\lim_{x \to 0^+} f(x) = 1$. Since these are unequal, $\lim_{x \to 0} f(x)$ does not exist. …

Why This Problem Matters

This is a classic "sign function" disguised in absolute-value clothing. The core idea is simple: continuity at a point means three things must match — the function value, the left-hand limit, and the right-hand limit. If any one of these is off, the function breaks at that point.

The trick here is that $|x|/x$ behaves differently depending on whether $x$ is positive or negative. For $x>0$, $|x| = x$, so the fraction is $1$. For $x<0$, $|x| = -x$, so the fraction is $-1$. That jump from $-1$ to $1$ at $x=0$ is the whole story.

Step-by-Step Solution

1. Understand the definition of $f$

The function is piecewise-defined:

• For $x \neq 0$: $f(x) = \frac{|x|}{x}$
• For $x = 0$: $f(0) = 0$

The absolute value $|x|$ is defined as:

2. Simplify $f(x)$ for $x \neq 0$

• If $x > 0$: $|x| = x$, so $f(x) = \frac{x}{x} = 1$.
• If $x < 0$: $|x| = -x$, so $f(x) = \frac{-x}{x} = -1$.

So the function is really:

This is the signum function (sign function) with a twist: at $x=0$, it's defined as $0$ instead of being undefined.

3. Check continuity at $x=0$ (the only potential trouble spot)

For any $x \neq 0$, the function is constant ($-1$ or $1$), so it's trivially continuous everywhere except possibly at $x=0$. We only need to examine $x=0$.

4. Compute the left-hand limit as $x \to 0^-$

When $x$ approaches $0$ from the left (negative side), $f(x) = -1$ for all such $x$. Therefore:

5. Compute the right-hand limit as $x \to 0^+$

When $x$ approaches $0$ from the right (positive side), $f(x) = 1$ for all such $x$. Therefore:

6. Compare the limits and the function value

We have:

• Left-hand limit: $-1$
• Right-hand limit: $1$
• Function value at $0$: $f(0) = 0$ …

Method: Rewriting an Absolute-Value Formula, Then Applying the Three-Condition Test

This method applies whenever a function's formula contains an expression like $|x|$ combined with $x$ itself (e.g. $\frac{|x|}{x}$ or $\frac{x}{|x|}$), since such a formula secretly hides a piecewise rule.

Steps

Step 1: Split the absolute value by the sign of $x$

Before doing anything else, rewrite $|x|$ using its own definition:

Substitute this into the given expression separately for $x>0$ and $x<0$ so the "hidden" piecewise structure becomes explicit.

Step 2: Identify the point where the rule changes

The sign of $x$ only switches at $x=0$, so this is the only candidate for a discontinuity; on either side of $0$ the simplified expression is a constant, which is automatically continuous.

Step 3: Apply the three-condition test at that point …

Common Mistakes

Mistake 1: Assuming $\frac{|x|}{x}$ simplifies to $1$ for every $x \neq 0$

Why it's wrong: this forgets that $|x| = -x$ when $x$ is negative, which flips the fraction to $-1$ on the left side of $0$ — the value is not the same on both sides. Correct approach: split by the sign of $x$ first ($x>0$ and $x<0$ separately), and simplify each branch on its own before drawing any conclusion.

Mistake 2: Plugging $x=0$ directly into $\frac{|x|}{x}$ to "find" $f(0)$ …