In such way some condition implies nonexistence of derivative? Assume that a function $f:[0,1] \rightarrow \mathbb{R}$ is continuous. In what way condition
$$\forall_{n\in \mathbb{N}} \forall_{0\leq x\leq 1-\frac{1}{n}} \exists_{0<h<1-x}  \left| \frac{f(x+h)-f(x)}{h}\right|>n$$
implies nonexistence of the finite right derivative of $f$ in each point from $[0,1)$ ?

Fix $x\in [0,1)$. For $n$ large enough (say $\geq N(x)$, we have $0\leq x\leq 1-\frac 1n$, and we can find for $n\geq N(x)$, $0<h_n<\frac 1n$ such that $\left|\frac{f(x+h_n)-f(x)}{h_n}\right|>n$. If the limit $\lim_{h\to 0^+}\frac{f(x+h)-f(x)}h$ existed (and was equal to $l$), then we would have  for $n$ large enough such that $\left|\frac{f(x+h_n)-f(x)}{h_n}-l\right|\leq 1$
$$n\leq \left|\frac{f(x+h_n)-f(x)}{h_n}\right|\leq \left|\frac{f(x+h_n)-f(x)}{h_n}-l\right|+|l|\leq l+1,$$
which is not possible.
