Let $f : \mathbb{R} \to \mathbb{R}$ be a Lipschitz function. Suppose $$ \lim_{n\to \infty} n[f(x + \frac{1}{n}) - f(x)] = 0.$$ Prove that $f$ is differentiable.
I am tempted to just let $n = \frac{1}{h},$ then $$\lim_{h\to 0} \frac{f(x+h) - f(x)}{h} = 0,$$
meaning that the limit exists and therefore $f$ is differentiable. However, I feel like the issue may be more subtle than this. Could someone give a more rigourus proof that might not involve using $n = \frac{1}{h}?$
Knowing the limit $\frac{f(x+h_n)-f(x)}{h_n} \to 0$ along a single sequence $h_n = \frac{1}{n}\to 0$ is not enough to be able to conclude that the limit along all possible sequences $h_n\to 0$ exists and is equal zero (which is needed to prove differentiabillity).
To solve the problem let $[h]$ denote the closest number to $h$ on the from $\frac{1}{n}$ with $n\in\mathbb{N}$ and write: $$\frac{f(x+h) - f(x)}{h} = \frac{f(x+h) - f(x+[h])}{h - [h]} \cdot \frac{h-[h]}{h} + \frac{f(x+[h]) - f(x)}{[h]}\cdot \frac{[h]}{h}$$ Now take $h\to 0$. The first term can be bounded using the Lipschitz condition, $\frac{|f(x)-f(y)|}{|x-y|}\leq C$, and the second term can be bounded using the given limit. You also need to show that $\frac{h-[h]}{h}\to 0$ and that $\frac{[h]}{h}$ is bounded as $h\to 0$ to reach the desired conclusion but this is fairly straight forward from the definition of $[h]$.