If $f$ is differentiable and its derivative is uniformly continuous, then $n \cdot (f(x+\frac{1}{n})-f(x)) \to^{u} f'(x)$ Let $f$ be a differentiable function in $\mathbb{R}$ such that its derivative $f'$ is uniformly continuous in $\mathbb{R}$. 
Prove: $$n \cdot (f(x+\frac{1}{n})-f(x)) \to^{u} f'(x)$$ (uniform convergence).
If $f'$ is only continuous- Does the claim still hold?
Please help me approach this one
 A: Let $\epsilon > 0$, so we have $\delta > 0$ such that $|x-y|< \delta$, $|f'(x) - f'(y)| < \epsilon$. Now let $n > \frac{1}\delta$ and $x \in \mathbb R$ $$\left|f\left(x + \frac1n\right) - f(x) - \frac1nf'(x)\right| = \left|\int_{0}^{\frac{1}n} (f'(x + t) - f'(x)) \mathrm dt\right| \le \frac1n\epsilon $$ Now you can prove uniform convergence. 
If $f'$ is only continuous this does not hold. For example $f(x) = \frac13x^3$ then $f'(x) = x^2$ and so $$n\left(f\left(n^2 + \frac1n\right) - f(n^2)\right) - f'(n^2) = \frac{n^2}{n^2} + \frac1{3n^3} \to 1$$ 
A: Instead of using integration, you can use the mean value theorem as suggested in the comments. 
Let $\epsilon > 0$, so we have $\delta > 0$ such that $|x-y|< \delta$, $|f'(x) - f'(y)| < \epsilon$. Now let $n > \frac{1}\delta$ and $x \in \mathbb R$. By the mean value theorem, $\exists \xi \in (x, x+\frac{1}{n})$ such that 
$$\frac{f\left(x+\frac{1}{n}\right) - f(x)}{\frac{1}{n}}=f'(\xi).$$
Since $n > \frac{1}{\delta}$, $|\xi - x| < \delta$, so $|f'(\xi) - f'(x)|<\epsilon$. Combining these two results,
$$\left|\frac{f\left(x+\frac{1}{n}\right) - f(x)}{\frac{1}{n}}-f'(x)\right|=\left|f'(\xi)-f'(x)\right|<\epsilon,$$
as required.
