Show LHL = RHL = f(0) = 0, so continuous. But not differentiable.

Check Continuity of f(x) = |x| at x = 0

Question

Check whether $f(x) = |x|$ is continuous at $x = 0$ .

Solution — Step by Step

For $f$ to be continuous at $x = 0$ , three conditions must hold simultaneously:

$f(0)$ exists
$\lim_{x \to 0} f(x)$ exists
$\lim_{x \to 0} f(x) = f(0)$

We’ll check each one.

By definition, $|0| = 0$ . So $f(0) = 0$ .

That’s the easy part.

As $x \to 0^-$ , $x$ is negative. For negative $x$ , we have $|x| = -x$ .

\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0

As $x \to 0^+$ , $x$ is positive. For positive $x$ , $|x| = x$ .

\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0

We have LHL = RHL = $f(0) = 0$ .

Since all three match, $f(x) = |x|$ is continuous at $x = 0$ .

Why This Works

The modulus function simply “folds” the negative side of the number line onto the positive side. Near $x = 0$ , the function approaches $0$ from both sides without any jump or gap — that’s exactly what continuity means geometrically.

The graph of $y = |x|$ is the classic V-shape. The vertex sits at the origin, and there’s no break in the curve there. This visual confirms what our algebra shows: the function is perfectly continuous at $x = 0$ .

Notice why we had to split the limit into LHL and RHL. The modulus function has different expressions on either side of $0$ , so it’s piecewise in nature. Any time the function definition changes at the point being tested, always compute both one-sided limits.

A standard follow-up in NCERT and CBSE board exams: “Is $f(x) = |x|$ differentiable at $x = 0$ ?” The answer is no. The left-hand derivative is $-1$ and the right-hand derivative is $+1$ — they don’t match. Continuous but not differentiable. This distinction is one of the highest-weightage conceptual points in the chapter.

Alternative Method

We can use the $\varepsilon$ - $\delta$ definition to confirm, but for board exams and JEE Main, the LHL = RHL = f(a) approach is faster and always accepted.

Another way to see it: rewrite $|x|$ using the identity $|x| = \sqrt{x^2}$ .

\lim_{x \to 0} \sqrt{x^2} = \sqrt{\lim_{x \to 0} x^2} = \sqrt{0} = 0

This works because $\sqrt{\cdot}$ is continuous, so we can pass the limit inside. It’s a neat one-liner, but the step-by-step LHL/RHL method is what examiners expect to see in written solutions.

Common Mistake

Many students write $\lim_{x \to 0} |x| = |0| = 0$ and call it done — but this skips the LHL and RHL verification. The examiner wants to see that you tested both sides separately. For $f(x) = |x|$ , the answer still comes out correct, but for questions like $f(x) = \frac{|x|}{x}$ (which is discontinuous at $0$ ), blindly substituting $x = 0$ will give you the wrong conclusion entirely. Always split the limit when the function involves $|x|$ near the test point.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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