Prove that $f_n(x)$ converges uniformly on $[a,b]$ Let $f(x)$ be continuously differentiable in $\mathbb{R}$, and let $f_n=n(f(x+1/n)-f(x))$.
Find $\displaystyle\lim_{n\to\infty}f_n(x)$, and prove that $(f_n(x))$ converges uniformly in $[a,b]$ for all $a,b\in\mathbb{R}$.
I already found that 
$$\lim_{n\to\infty}f_n(x)=f'(x)$$ but I have no idea how to prove the second part. I'd be grateful if someone could give me a hint :)
 A: Hints: use the Mean Value Theorem (on $f$) and (uniform) continuity (of $f'$).
(Spoilers below.)

In detail:
So you have shown that $(f_n)_n$ converges pointwise to $f'$.
Now, for any $x\in[a,b]$, writing $h\stackrel{\rm def}{=} \frac{1}{n}$,
$$
\lvert f_n(x) - f'(x) \rvert
= \left\lvert \frac{f(x+h_n)-f(x)}{h_n} - f'(x) \right\rvert
= \left\lvert f'(c_n) - f'(x) \right\rvert \tag{1}
$$
for some $c_n\in(x,x+h_n)$, by the Mean Value Theorem. That's where we use the fact that $f'$ is continuous on $[1,b]$, and therefore uniformly continuous...
Fix $\varepsilon > 0$. There exists $\delta > 0$ such that, for any $y,x\in[a,b]$ such that $\lvert x-y\rvert \leq \delta$, we have $\left\lvert f'(c_n) - f'(x) \right\rvert\leq \varepsilon$. In particular, we let $n_\varepsilon\geq 1$ be such that $\frac{1}{n_\varepsilon} \leq \delta$. For any $n\geq n_0$, we then have that, by (1)
$$\forall x\in[a,b], \quad
\left\lvert f'(c_n) - f'(x) \right\rvert \leq \varepsilon
$$
since $\lvert c_n - x\rvert \leq h_n = \frac{1}{n} \leq \delta$.
In summary: for any $\varepsilon > 0$, there exists $n_\varepsilon\geq 1$ such that for any $n\geq n_\varepsilon$,
$$
\sup_{x\in[a,b]} \lvert f_n(x) - f'(x) \rvert \leq \varepsilon
$$
which shows uniform convergence.
A: Let $\varepsilon>0$. We want to find $n_0$ such that for each $n\ge n_0$ and for each $x\in\left[a,b\right]$, $\left|f_n(x)-f'(x)\right|<\varepsilon$. That is,
$$\left|\frac{f\left(x+\frac{1}{n}\right)-f\left(x\right)}{\frac{1}{n}}-f'(x)\right|<\varepsilon$$
By Lagrange's theorem, there is some $c\in\left[x,x+\frac{1}{n}\right]$ such that $f'(c)=\frac{f\left(x+\frac{1}{n}\right)-f\left(x\right)}{\frac{1}{n}}$. So the above inequality is $\left|f'(c)-f'(x)\right|<\varepsilon$.
$f'$ is continuous on $\left[a,b\right]$, hence uniformly continuous. So there is some $\delta>0$ such that if $x_1,x_2\in\left[a,b\right]$ satisfy $\left|x_2-x_1\right|<\delta$, we have $\left|f'(x_1)-f'(x_2)\right|<\varepsilon$.
Taking some $n_0$ such that $\frac{1}{n_0}<\delta$ and combining all of the above, we get the desired conclusion.
