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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 $\lim_{x \to 0} g (x) = 1$.)

DDS
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  • It is not valid as is. More justification is required to use that argument. – Rushabh Mehta Jul 20 '19 at 15:35
  • See https://math.stackexchange.com/questions/97213/how-can-a-function-have-a-vertical-tangent-and-be-continuous https://math.stackexchange.com/questions/2356839/can-you-have-a-vertical-tangent-where-a-function-is-undefined – user600016 Jul 20 '19 at 15:36

3 Answers3

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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 $x \rightarrow 0$.

Aphelli
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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 $$\lim_{x\to c}f'(x)=f'(c)$$

Prof. Shanku
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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.

DDS
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