Let $f:\mathbb R\to\mathbb R$ be a function and suppose that for all $n\in\mathbb N$, there exist some $a_n,b_n\in\mathbb R$ such that $$0\leq a_n
I want to show that $f$ is not differentiable at $x=0$ in this case. Of course, the non-existence of a right-hand derivative would be sufficient, yet I seem to be stuck with proving this.
Any hints (or counterexamples, for that matter) would be greatly appreciated.
On a second thought, I could not prove this conjecture because it is false. Counterexample: \begin{align*} f(x)=\begin{cases}x^2&\text{if $x$ is rational,}\\0&\text{if $x$ is irrational.}\end{cases} \end{align*} This function is differentiable at $x=0$, yet it admits arbitrarily large slopes near $0$.
What happens if $f$ is also required to be continuous everywhere, though? If $f$ is also required to be continuous, then the following function will do as a counterexample:
\begin{align*}
f(x)=\begin{cases}x^2\sin(x^{-2})&\text{if $x\neq0$,}\\0&\text{if $x=0$.}\end{cases}
\end{align*}
This function is easily checked to be differentiable, even at $0$, with
\begin{align*}
f'(0)=&\;0,\\
f'\left(\frac{1}{\sqrt{k\pi}}\right)=&\;2\sqrt{k\pi}\quad\text{if $k\in\mathbb N$ is odd.}
\end{align*}
Therefore, $f'(x)$ can be made arbitrarily large for arbitrarily small $x>0$, and thus so can the corresponding slopes via approximation.