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Differentiability at a cusp using the limit definition

Consider the function f(x)=∣x∣f(x) = \sqrt{|x|}. Its graph has a sharp point at the origin, often called a cusp.

Use the limit definition of the derivative, f′(0)=lim⁡h→0f(0+h)−f(0)h,f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}, to determine whether f(x)=∣x∣f(x) = \sqrt{|x|} is differentiable at x=0x = 0. Evaluate the one-sided limits as h→0+h \to 0^+ and h→0−h \to 0^- separately, then state whether f′(0)f'(0) exists.

Differentiability at a cusp using the limit definition graph
Hint

Write out f(h)−f(0)h=∣h∣h\frac{f(h)-f(0)}{h} = \frac{\sqrt{|h|}}{h} and simplify separately for h>0h>0 (where ∣h∣=h|h|=h) and h<0h<0 (where ∣h∣=−h|h|=-h). Compare the two resulting limits as h→0h \to 0.

Step-by-step solution

For h>0h > 0, ∣h∣=h|h| = h, so the difference quotient is hh=1h\frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}}. As h→0+h \to 0^+, 1h→+∞\frac{1}{\sqrt{h}} \to +\infty, so this one-sided limit does not exist (it diverges).

For h<0h < 0, ∣h∣=−h|h| = -h, so the difference quotient is −hh=−1−h\frac{\sqrt{-h}}{h} = -\frac{1}{\sqrt{-h}}. As h→0−h \to 0^-, −1−h→−∞-\frac{1}{\sqrt{-h}} \to -\infty. Since the two one-sided limits both diverge (to opposite infinities), f′(0)=lim⁡h→0∣h∣hf'(0) = \lim_{h\to 0} \frac{\sqrt{|h|}}{h} does not exist, and the graph shows a vertical tangent-like cusp at the origin, so ff is not differentiable at x=0x=0.

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