Continuity & Differentiability: Common Mistakes and Fixes (7)

easy • 2 min read
Tags Continuity And Differentiability

Question

Examine the continuity and differentiability of f(x)=∣x−2∣f(x) = |x - 2| at x=2x = 2.

Solution — Step by Step

f(x)=∣x−2∣f(x) = |x - 2| means f(x)=x−2f(x) = x - 2 if x≥2x \geq 2, and f(x)=2−xf(x) = 2 - x if x<2x < 2.

Left limit at x=2x = 2: lim⁡x→2−(2−x)=0\lim_{x\to 2^-}(2 - x) = 0. Right limit at x=2x = 2: lim⁡x→2+(x−2)=0\lim_{x\to 2^+}(x - 2) = 0. f(2)=0f(2) = 0.

All three match, so ff is continuous at x=2x = 2.

 f′(2−)=lim⁡h→0−f(2+h)−f(2)h=lim⁡h→0−∣2+h−2∣−0h=lim⁡h→0−∣h∣h=−1f'(2^-) = \lim_{h \to 0^-}\frac{f(2 + h) - f(2)}{h} = \lim_{h\to 0^-}\frac{|2 + h - 2| - 0}{h} = \lim_{h\to 0^-}\frac{|h|}{h} = -1

(Since h<0h < 0, ∣h∣=−h|h| = -h, so the ratio is −1-1.)

 f′(2+)=lim⁡h→0+∣h∣h=1f'(2^+) = \lim_{h\to 0^+}\frac{|h|}{h} = 1

f′(2−)=−1≠1=f′(2+)f'(2^-) = -1 \ne 1 = f'(2^+). The two one-sided derivatives differ, so ff is not differentiable at x=2x = 2.

ff is continuous at x=2x = 2 but not differentiable there.

Why This Works

Continuity requires the function value to match the two-sided limit. Differentiability is stronger: it requires the slope to be the same from both sides. The graph of ∣x−2∣|x - 2| has a sharp corner at x=2x = 2 — the slopes are −1-1 on the left and +1+1 on the right, so no single tangent exists.

This is the canonical example of a function that is continuous but not differentiable. Every absolute-value function has this property at its kink.

Alternative Method

Geometric argument: ∣x−2∣|x - 2| has a V-shaped graph with vertex at x=2x = 2. The left arm has slope −1-1, the right arm slope +1+1. A tangent line cannot be defined at a vertex, so the function is not differentiable there.

Common Mistake

Concluding non-continuity because the function “changes definition” at x=2x = 2. Continuity is about limits matching the value, not about the formula being uniform. Many functions are continuous despite being defined piecewise.

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