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Differentiability of $x^{1/3}$ at $x=0$ via limit definition

Consider the function f(x)=x1/3f(x) = x^{1/3}, whose graph has a vertical tangent line at x=0x=0.

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)=x1/3f(x) = x^{1/3} is differentiable at x=0x = 0. Compute the limit (or show it does not exist as a finite number), and state whether ff is differentiable at x=0x=0.

Differentiability of $x^{1/3}$ at $x=0$ via limit definition graph
Hint

Substitute f(0)=0f(0)=0 and f(h)=h1/3f(h) = h^{1/3} into the difference quotient, simplify to h−2/3h^{-2/3}, and think about what happens to this expression as h→0h \to 0 from both sides.

Step-by-step solution

Plug into the definition: f′(0)=lim⁡h→0h1/3−0h=lim⁡h→0h1/3h=lim⁡h→0h−2/3f'(0) = \lim_{h\to 0} \dfrac{h^{1/3} - 0}{h} = \lim_{h\to 0} \dfrac{h^{1/3}}{h} = \lim_{h\to 0} h^{-2/3}. This simplifies because h1/3/h=h1/3−1=h−2/3h^{1/3}/h = h^{1/3-1} = h^{-2/3}.

As h→0h \to 0 from either side, h−2/3=1/h2/3→∞h^{-2/3} = 1/h^{2/3} \to \infty, since the denominator h2/3h^{2/3} is positive and approaches 00. So the limit does not exist as a finite number, meaning f′(0)f'(0) does not exist and ff is not differentiable at x=0x=0. This corresponds to the graph having a vertical tangent line at that point.

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