uniform convergent Given f is a differentiable function. Define $$f_n(x)=n\left(f\left(x+\frac{1}{n}\right)-f(x)\right)$$, prove that $f_n$ is uniformly converge to $f'$
I have tried to make these equations:
$n(f(x+\frac{1}{n})-f(x))=\frac{f(x+\frac{1}{n})-f(x)}{1/n}$ and taking limits as n $\to\infty$ but i got stuck to prove its uniform convergent

This is true only when $f'$ is uniformly continuous.(Use MVT) 
For, 
By MVT, there exists $\varepsilon_n \in (x,x+1/n)$ such that $$\Big|\frac{f(x+1/n)-f(x)}{1/n}-f'(x)\Big|=|f'(\varepsilon_n)-f'(x)|$$
How to make the RHS small in order to make uniform convergent?
Ans: Using uniform continuity of $f'$
Otherwise, consider, for example, $f(x)=x^3$ on $\Bbb R$. Then $$\sup_x \Big|\frac{f(x+1/n)-f(x)}{1/n}-f'(x)\Big|=\sup_\color{red}x \Big|\frac{3x}{n}+\frac{1}{n^2}\Big|= \infty,$$concluding the convergence is not uniform!
