Q.Is the function defined by $f(x) = |x|$, a continuous function?

The key idea is continuity at a point: a function is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. For $f(x) = |x|$, we check this at every real number.

1. For any $a > 0$, near $a$ we have $|x| = x$, so $\lim_{x \to a} |x| = a = f(a)$.
2. For any $a < 0$, near $a$ we have $|x| = -x$, so $\lim_{x \to a} |x| = -a = f(a)$. …

Why This Question Matters

Many students see the V-shaped graph of $|x|$ with its sharp point at $x = 0$ and instinctively think "that's not continuous — it has a corner." But continuity and differentiability are different ideas. A function can be continuous at a point even if it has a corner, a cusp, or a vertical tangent. The only thing continuity demands is that the graph doesn't break — you should be able to trace it without lifting your pen.

The absolute value function is the classic example that separates these two concepts. Let's prove it properly.

Step-by-Step Proof

1. Recall the definition of continuity at a point

A function $f$ is continuous at $x = a$ if and only if three conditions hold:

• $f(a)$ is defined
• $\lim_{x \to a} f(x)$ exists
• $\lim_{x \to a} f(x) = f(a)$

For $f(x) = |x|$, we need to check this at every real number $a$. The only potentially tricky point is $a = 0$, because that's where the definition of $|x|$ changes.

2. First, handle all points where $a \neq 0$

If $a > 0$, then near $a$ we have $f(x) = x$ (since $x$ stays positive in a small enough neighbourhood). The function $g(x) = x$ is a polynomial, and polynomials are continuous everywhere. So $\lim_{x \to a} |x| = a = |a|$.

If $a < 0$, then near $a$ we have $f(x) = -x$ (since $x$ stays negative nearby). The function $h(x) = -x$ is also a polynomial, hence continuous. So $\lim_{x \to a} |x| = -a = |a|$.

3. Now check $x = 0$ — the critical point

Here $f(0) = |0| = 0$, so condition 1 is satisfied.

For the limit, we need to check both sides separately because the function's rule changes at 0.

Left-hand limit: As $x \to 0^-$, $x$ is negative, so $|x| = -x$. Therefore:

Right-hand limit: As $x \to 0^+$, $x$ is positive, so $|x| = x$. Therefore:

Since both one-sided limits equal 0, the two-sided limit exists and: …

Method: Checking Global Continuity of a Piecewise-Defined Function

This method decides whether a function built from more than one formula (like an absolute value) is continuous everywhere, by splitting the real line into the "safe" regions where a single simple formula applies and the "boundary" point(s) where the formula changes.

Steps

Step 1: Identify where the function's rule changes

Locate the value(s) of $x$ at which the piecewise definition switches from one formula to another — these are the only points that need special attention.

Step 2: Argue continuity on each open region away from the boundary

On any interval that does not contain a switching point, the function equals a single simple formula (e.g., a polynomial like $x$ or $-x$). Since such elementary functions are already known to be continuous everywhere, the function is automatically continuous throughout each such region — no separate limit computation is needed.

Step 3: Check continuity at each switching point individually …

Common Mistakes

Mistake 1: Only checking $x=0$ and forgetting the rest of the domain

Why it's wrong: the question asks whether $f(x)=|x|$ is continuous as a function (everywhere), not just at one point — checking only the corner at $x=0$ and ignoring all other $x$ leaves the proof incomplete. Correct approach: argue continuity at $a \neq 0$ using the fact that $|x|$ equals a simple polynomial ($x$ or $-x$) on either side, and handle $x=0$ separately with one-sided limits.

Mistake 2: Assuming the corner at $x=0$ automatically breaks continuity …