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Limit definition of the derivative at a corner/cusp point

Consider the function f(x)=x2/3f(x) = x^{2/3}. We want to investigate whether this function has a well-defined tangent line at x=0x = 0 by examining the difference quotient there.

Using 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}, compute this limit for f(x)=x2/3f(x) = x^{2/3} at x=0x=0, and use your result to determine whether ff is differentiable at x=0x = 0.

Limit definition of the derivative at a corner/cusp point graph
Hint

Write out h2/3−0h\frac{h^{2/3} - 0}{h} and simplify using exponent rules, noting that h2/3=h−1/3⋅hh^{2/3} = h^{-1/3} \cdot h. Then think about what happens to h−1/3h^{-1/3} as h→0+h \to 0^+ versus h→0−h \to 0^-.

Step-by-step solution

We form the difference quotient: f(0+h)−f(0)h=h2/3−0h=h2/3h\dfrac{f(0+h)-f(0)}{h} = \dfrac{h^{2/3} - 0}{h} = \dfrac{h^{2/3}}{h}. Simplifying the exponents, h2/3h1=h2/3−1=h−1/3=1h1/3\dfrac{h^{2/3}}{h^1} = h^{2/3 - 1} = h^{-1/3} = \dfrac{1}{h^{1/3}}.

As h→0+h \to 0^+, h1/3→0+h^{1/3} \to 0^+, so 1h1/3→+∞\dfrac{1}{h^{1/3}} \to +\infty. As h→0−h \to 0^-, h1/3→0−h^{1/3} \to 0^-, so 1h1/3→−∞\dfrac{1}{h^{1/3}} \to -\infty. Since the limit does not approach a single finite value, lim⁡h→0h−1/3\lim_{h\to 0} h^{-1/3} does not exist, so f′(0)f'(0) does not exist. Geometrically, this corresponds to a vertical tangent line at x=0x=0, and ff is not differentiable there (even though it is continuous there).

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