Prove that $f: \mathbb{R} \to \mathbb{R}, f(x) = x^\frac{1}{9}$ is not a differentiable function Prove that the following function is not a differentiable function: $$f: \mathbb{R} \to \mathbb{R}, f(x) = x^\frac{1}{9}$$
I believe all I have to show is one point in the domain where the function is not differentiable:
Hence I have the following proof:
$$ f'(x) =\frac{1}{9x^\frac{8}{9}} $$
At $ x = 0$, $f'(x)$ is undefined.
Hence, $f(x)$ is not a differentiable function.
Is this enough or do I need to show that the $\lim\limits_{h\to0} \frac{f(x_0 + h) - x_0}{h}$ does not exist in some other way.
 A: Yes, you must show that the limit $\lim_{x\to 0}\frac{f(x)-f(0)}{x}$does not exist. 
Note that $f'(x)$ exists for all $x\ne 0$ however at $x=0$, differentiability is doubtful that is to say it is apriori not known whether $f'(0)$ exists or not and if it does exist then what form it will be in. 
What you have shown above is that $f'(x) =\frac{1}{9x^\frac{8}{9}}$ does not exist at $x=0$, which has nothing to do with $f'(0)$ as we don't even know apriori about existence of $f'(0)$ or the kind of form $f'(0)$ will be in if it exists.
Therefore, by definition of differentiability at $x=0$, $f'(0)$ exists if and only if $\lim_{x\to 0}\frac{f(x)-f(0)}{x}$ exists. 
Clearly, $\frac{f(x)-f(0)}{x}=\frac{x^{1/9}-0}{x}=\frac 1{x^\frac 89}$ (Note that it is different from $\frac{1}{9x^\frac{8}{9}}$) and therefore the limit does not exist finitely, which is to say that $f'(0)$ is non-existent.
A: As others pointed out, the non-existence of $\lim_{x\to 0}f'(x)$ does not imply that $f$ is not differentiable at $x=0$. However, the following is correct and can be applied here:

Let $f: (a-h, a+h) \to \Bbb R$ be continuous, and differentiable on $(a-h, a+h) \setminus \{ a \}$. If $\lim_{x\to a}|f'(x)| = \infty$ then $f$ is not differentiable at $x=a$.

Proof: This is a consequence of the mean-value theorem: Assume that $f'(a)$ exists, and let $x_n$ be a sequence in  $(a-h, a+h) \setminus \{ a \}$ which converges to $a$. Then
$$
 \frac{f(x_n)-f(a)}{x_n-a} = f'(c_n)
$$
for some $c_n$ between $a$ and $x_n$. The left-hand side converges to $f'(a)$, whereas the right-hand side is unbounded for $n \to \infty$. This is a contradiction.

In your case, $f(x) = x^{1/9}$ is continuous on $\Bbb R$, differentiable on $\Bbb R \setminus \{ 0\}$, and
$$
 \lim_{x\to 0}|f'(x)| = \lim_{x\to 0}\frac{1}{9|x|^{8/9}} = \infty \, ,
$$
which implies that $f$ is not differentiable at $x=0$.
A: Another idea is to use sequence to prove that, suppose $a(n)=\frac {(-1)^n}{n}$ for big $n$ we have  $\lim_{n\to \infty}a(n)\to 0 $ so we can use $a(n),n\to \infty$ instead of $x\to 0$
now if you put $a(n)$ as $x$ $$\lim_{x \to 0}\frac{f(x)-f(0)}{x-0} =\lim_{n \to \infty}\frac{f(a(n))-f(0)}{a(n)-0}=\\
\lim_{n \to \infty}\frac{f(\frac {(-1)^n}{n})-f(0)}{\frac {(-1)^n}{n}-0}=\\
\lim_{n \to \infty}\frac{f(\frac {(-1)^n}{n})}{\frac {(-1)^n}{n}}=\\
\lim_{n \to \infty}(-1)^nn\sqrt[9]{\frac {(-1)^n}{n}}=\\
\lim_{n \to \infty}\sqrt[9]{\frac {(-1)^n\times n^9\times (-1)^{9n}}{n}}=\\
\lim_{n \to \infty}\sqrt[9]{\frac { n^8\times (-1)^{10n}}{1}}=\\
\lim_{n \to \infty}\sqrt[9]{\frac { n^8\times (+1)}{1}}\to +\infty\\$$
A: I quite do not agree with the other answer. Computing $f'(x)$ for $x\neq 0$ and saying it does not have a limit at $0$ does not show that $f'(0)$ does not exist: it shows that if $f'$ is defined at $0$, it cannot be continuous at $0$. However, a short adaptation of this gives the answer: for $x\neq 0$,
$$
\frac{f(x)-f(0)}{x-0} = \frac{x^{1/9}}{x} = x^{-8/9}
$$
which does not have a limit at $0$. This shows that $f$ is not differentiable at $0$.
Edit As suggested in comment, here is an example of a differentiable function $f : \mathbb{R} \to \mathbb{R}$ such that $f'$ does not have a limit at $0$ while $f'(0)$ does exist. Define
\begin{align}
f : \mathbb{R} & \longrightarrow \mathbb{R} \\
x & \longmapsto \left\{\begin{array}{rcl} x^2 \sin(1/x) & \text{if} & x \neq 0 \\
0 & \text{if} & x=0
\end{array}\right.
\end{align}
Then $f$ is differentiable on $\mathbb{R} \setminus\{0\}$ by easy calculations, where $f'(x) = 2x\sin(1/x) - \cos(1/x)$. This expression does not have a limit when $x$ goes near $0$. But if $x \neq 0$,
$$
\frac{f(x) - f(0)}{x-0} = \frac{x^2\sin(1/x)}{x} = x \sin(1/x) \underset{x\to 0}{\longrightarrow} 0
$$
which shows that $f'(0)$ exists and is equal to $0$.
