Exercise 398Calculus › Single-variable calculus › Differentiation › Differentiable functionsL1Added 2 Nov 2026

Three ways to fail to be differentiable

#differentiability#corner#vertical tangent#oscillation

Statement

Each of the following functions is continuous at $0$ . Show that none is differentiable at $0$ , and explain why in each case.

$f_1(x)=|x|$
$f_2(x)=\sqrt[3]{x}$
$f_3(x)=x\sin\dfrac1x$ for $x\ne0$ , $f_3(0)=0$

Hint

Write the difference quotient $\dfrac{f(h)-f(0)}h$ in each case and study its behaviour as $h\to0$ .

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Solution

1. A corner. The quotient is $\frac{|h|}h$ , equal to $1$ for $h>0$ and $-1$ for $h<0$ . The two one-sided derivatives exist but differ.

2. A vertical tangent. The quotient is $\dfrac{h^{1/3}}h=h^{-2/3}\to+\infty$ as $h\to0$ (from both sides). The slope of the secants becomes infinite.

3. Wild oscillations. The quotient is $\dfrac{h\sin(1/h)}h=\sin\dfrac1h$ , which has no limit as $h\to0$ : it takes the values $0$ and $1$ along suitable sequences $h\to0$ . The secants through $(0,0)$ keep oscillating in slope between $-1$ and $1$ .

Moral. Differentiability is a strictly stronger property than continuity, and it can fail in three different ways: left and right slopes differ, the slope blows up, or the slope has no limit at all.

Three ways to fail to be differentiable

Statement

Solution

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