Continuity and differentiability of |x| at x = 0 — graphical explanation

Question

Show that $f(x) = |x|$ is continuous but not differentiable at $x = 0$ . Explain graphically.

(NCERT Class 12, Chapter 5 — Continuity and Differentiability)

Solution — Step by Step

f(x) = |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

Three conditions must hold:

$f(0) = |0| = 0$ ✓ (function is defined)
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$ ✓
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0$ ✓

Since left limit = right limit = $f(0) = 0$ , the function is continuous at $x = 0$ .

The derivative requires $\lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{|h|}{h}$ to exist.

Right-hand derivative: $\lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$

Left-hand derivative: $\lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$

Since $1 \neq -1$ , the derivative does not exist at $x = 0$ . The function is not differentiable at $x = 0$ .

The graph of $|x|$ is a V-shape with a sharp corner at the origin. The left arm has slope $-1$ and the right arm has slope $+1$ .

At the corner, there’s no unique tangent line — you could draw infinitely many lines through the point that “touch” the curve. Differentiability requires a unique tangent, which means a smooth curve with no sharp corners. The sharp point at $x = 0$ is exactly where differentiability fails.

Why This Works

Continuity means “no breaks” — the graph can be drawn without lifting the pen. Differentiability means “no sharp corners” — the graph must be smooth enough to have a unique tangent at every point.

A function can be continuous without being differentiable (like $|x|$ at the corner), but a differentiable function is always continuous. So differentiability is a “stronger” condition than continuity.

The derivative is the slope of the tangent line. At the corner of $|x|$ , the slope changes abruptly from $-1$ to $+1$ — there’s no single slope value that works, so the derivative doesn’t exist.

Alternative Method — Using the formal definition with specific sequences

Take $h_n = 1/n$ (approaching 0 from the right): $\frac{|1/n|}{1/n} = 1$ for all $n$ .

Take $h_n = -1/n$ (approaching 0 from the left): $\frac{|-1/n|}{-1/n} = \frac{1/n}{-1/n} = -1$ for all $n$ .

Different sequences give different limits, so the limit doesn’t exist.

This is a classic example that separates continuity from differentiability. JEE loves asking: “Is $f(x) = |x - a| + |x - b|$ differentiable everywhere?” The answer: not at $x = a$ and $x = b$ (sharp corners). Any function involving absolute values likely has non-differentiable points — check left and right derivatives at each critical point.

Common Mistake

Some students conclude that since $f(x) = |x|$ has a derivative of $-1$ on the left and $+1$ on the right, the derivative at $x = 0$ must be the average: $0$ . This is wrong. The derivative is defined as a limit, and if the left and right limits disagree, the limit simply does not exist. Averaging is not a valid operation for limits that disagree.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the formal definition with specific sequences

Common Mistake

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