Let $f\colon\mathbb{R}\rightarrow\mathbb{R}$ be continuous and differentiable, where $f'(x)$ is uniformly continuous. Then prove that the following is true for all $x \in \mathbb{R}$ $$\lim_{n \rightarrow\infty}n\left(f\left(x+\frac{1}{n}\right)-f(x)\right)=f'(x).$$
My attempt:
Let $\varepsilon >0$ be arbitrary. Since $f'(x)$ is uniformly continuous, there exists $\delta$ such that if $|h| \leq \delta$, we have for any $x \in \mathbb{R}$ $$\frac{f(x+h)-f(x)}{h} \leq \varepsilon. $$
Consider the case where $h = \frac{1}{n}$. Then we can fix a $N \in \mathbb{N}$ such that $\frac{1}{N} \leq \delta$ and for any natural number $n \geq N$, we maintain the inequality above.
And now we can see that the equality is valid for any $n \geq N$: $$\lim_{n \rightarrow\infty}n\left(f\left(x+\frac{1}{n}\right)-f(x)\right)= \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} =f'(x).$$
that is, both limits converge since $f$ is differentiable and $f'(x)$ is uniformly continuous.
Note: I am not sure this is correct. I've read some hints which recommend using MVT, but I found no explanation on why one must use MVT nor how to use it.
Lets define $n^{'}$ := $\frac{1}{n} $ Notice as $ n \to \infty$ we have $n^{'} \to 0 $
Then $\lim_{n \rightarrow\infty}n\left(f\left(x+\frac{1}{n}\right)-f(x)\right) = \lim_{n^{'} \to 0} \frac{f(x+n^{'})-f(x)}{n^{'}} = f'(x)$