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 $[0,1)$ ?
 A: 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.
A: Consider the negation:
$$\exists_{n\in \mathbb{N}} \exists_{0\leq x\leq 1-\frac{1}{n}} \forall_{0<h<1-x}  \left| \frac{f(x+h)-f(x)}{h}\right|\leq n$$
Note that the negation is equivalent to the following property. (You might want to see if you can correctly write out all the details in a proof of their equivalence.)
$$\exists_{m\in \mathbb{N}} \exists_{n\in \mathbb{N}}  \exists_{0\leq x\leq 1-\frac{1}{m}} \forall_{0<h<1-x}  \left| \frac{f(x+h)-f(x)}{h}\right|\leq n$$
This last property should now be recognizable as the property of having bounded right difference quotients based at some $x \in [0,1).$ The parameter $m$ allows for all possible points $x \in [0,1)$ to be included and the parameter $n$ allows for all possible (finite) bounds to be included.
Thus, the property you gave is equivalent to: At each $x \in [0,1)$ and for each right neighborhood of $x$ (where the right neighborhood belongs to the interval $[0,1]$), the function $f$ does not have bounded difference quotients where the left point is fixed at $x$ and the right point varies over points in that right neighborhood. An equivalent way of stating this is to say that, for each $x \in [0,1)$, at least one of the two right Dini derivates of $f$ at $x$ (upper right derivate or lower right derivate) is not finite. Part of the difficulty you're having might be due to the fact that this is quite a bit stronger than saying that, for each $x \in [0,1),$ the function $f$ does not have a finite right derivative at $x.$
