# On the Differentiability of $f(x) = \sqrt[3]x$ at $x=0$

Suppose one is asked to explain why the real valued function

$$f(x) = \sqrt[3]{x}$$

is not differentiable at $$x = 0$$ and gives the following argument: For $$x \neq 0$$, one has

$$f'(x) = \frac{1}{3\sqrt[3]{x^2}}$$

and this is not defined for $$x = 0$$. Is this a perfectly valid argument? In my opinion, it is not since $$f'(0)$$ is a limit and it $$\textit{could}$$ be that the limit nevertheless exists. I mean the following: If one says that $$f'(0)$$ does not exists because $$\frac{1}{3 \cdot 0}$$ is not defined, one implicitly assumes that $$f'$$ is continuous at $$x = 0$$, or? I think, that the correct answer would be, that

$$\lim_{x \to 0} f'(x) = \infty.$$

(My question arises from the point that a function could have a limit in a point, although it is not defined there. Suppose for example you have $$g(x) = \frac{x}{x}$$. Then $$g$$ is not defined at $$x = 0$$ but $$\lim_{x \to 0} g (x) = 1$$.)

• It is not valid as is. More justification is required to use that argument. – Don Thousand Jul 20 at 15:35
• – thewitness Jul 20 at 15:36

That the expression giving $$f’(x)$$ for $$x > 0$$ is undefined at $$x=0$$ is not a valid reason to state that $$f$$ isn’t differentiable at $$0$$.

You can think for instance of $$f(x)=x^{3/2}\sin{\frac{1}{x}}$$: for $$x > 0$$, $$f’(x)=-x^{-1/2}\cos{\frac{1}{x}}+3/2x^{1/2}\sin{\frac{1}{x}}$$, so it is undefined at $$x=0$$. However, $$f’(0)=0$$.

On the other hand, if $$f’(x) \rightarrow \infty$$ when $$x \rightarrow 0$$, then $$f$$ isn’t differentiable at $$0$$. Indeed, for all $$x$$, $$\frac{f(x)-f(0)}{x}=f’(c_x)$$ for some $$0 < c_x < x$$, thus $$f’(c_x) \rightarrow \infty$$ as $$x \rightarrow 0$$.

For checking differentiability of a function,say $$f$$ at a point $$x=a$$ one should always use the definition of differentiability i.e. try to check if $$\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$ exists or not. It may so happen that the limit exists but the derivative function $$f'$$ is not continuous or defined at that particular point (Consider the classical example of the function: $$f(x)=x^2.sin(x)$$ when $$x\neq0$$ and $$f(x)=0$$ when $$x=0$$).

Important thing to note that if you have a function defined in the interval $$(a,b)$$ and suppose that $$c\in(a,b)$$ and $$f$$ is differentiable in $$(a,c)$$ and in $$(c,b)$$ and if $$\lim_{x\to c}f'(x)$$ exists then $$f$$ is indeed differentiable at $$c$$ and $$\lim_{x\to c}f'(x)=f'(c)$$

As suggested, start with the definition of the derivative at $$a = 0$$:

$$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x -0} = \lim_{x \to 0} \frac{\sqrt[3]x - 0}{x} = \lim_{x \to 0} \frac{1}{\sqrt[3]{x^{2}}}.$$

Hence, as both the LHL and RHL as $$x \to 0$$ is $$\infty$$ (i.e., D.N.E.), we conclude that there is an infinitely-sloped tangent line to the graph of $$f(x)$$ at $$x=0$$.

Therefore, $$f'(0)$$ does not exist.