Show that f(x)=|x| is continuous but not differentiable at x=0

Question

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

Solution — Step by Step

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

This piecewise form is essential — it lets us check left-hand and right-hand limits separately.

For $f(x)$ to be continuous at $x = 0$ , we need:

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

Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0$

Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x) = 0$

Value at x = 0: $f(0) = |0| = 0$

Since LHL = RHL = f(0) = 0, $f(x) = |x|$ is continuous at $x = 0$ . ✓

For differentiability, we need:

\lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \text{ to exist}

This means the left-hand derivative (LHD) must equal the right-hand derivative (RHD).

RHD (h → 0⁺):

\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{|h| - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1

LHD (h → 0⁻):

\lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{|h| - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1

Since LHD = $-1 \neq 1$ = RHD, the derivative does not exist at $x = 0$ .

$f(x) = |x|$ is NOT differentiable at $x = 0$ .

This is reflected geometrically: the graph of $|x|$ has a sharp corner (or “kink”) at the origin. A function is differentiable where its graph is smooth — at a corner, the tangent line is not uniquely defined.

Why This Works

Continuity asks: “Does the function value agree with the limit?” — it’s about the y-values matching.

Differentiability asks a sharper question: “Does the function have a well-defined tangent slope?” — it’s about the rate of change approaching from both sides.

For $|x|$ , approaching from the right the slope is +1 (the function is $y = x$ ), but from the left the slope is -1 (the function is $y = -x$ ). These disagree at the corner, so no derivative exists.

The key theorem: Differentiability implies continuity, but continuity does NOT imply differentiability. $|x|$ at $x = 0$ is the classic example of this gap.

Common Mistake

Students often stop at showing continuity and forget the second part (non-differentiability). Or they compute $f'(x) = \text{sgn}(x)$ and say “derivative at 0 is undefined” without showing the LHD/RHD mismatch. For CBSE Class 12 board exams, the examiner wants to see both LHD and RHD calculated explicitly, with the conclusion that LHD ≠ RHD.

Question

Solution — Step by Step

Why This Works

Common Mistake

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