Differentiable and continuous function related to a limit of a function Let $f$ be a differentiable function on the interval $(-2,2)$ such that $f'$ is continuous on this interval. Prove that $$\lim_{h\rightarrow 0}\int_{0}^{1}{[\frac{f(x+h)-f(x)}{h}-f'(x)]dx=0}$$
I need to find the $\lim_{h\rightarrow 0}g(h)=0$ where $g(h)=\int_{0}^{1}{[\frac{f(x+h)-f(x)}{h}-f'(x)]dx}$
Since $f$ is differentiable, $\exists \delta>0$ such that if $0<|h|<\delta$ then $|\frac{f(x+h)-f(x)}{h}-f'(x)|<\epsilon.$
But I do not know how to find the suitable $\delta$ such that $|g(h)|<\epsilon$. It would be great if someone can help me on this.
 A: Differential Calculus
Since $f'$ is continuous on $(-2,2)$, then function
$$
F(x) := \int_0^x f(t)\mathrm dt 
$$
is defined and twice continuously differentiable on $(-2,2)$ with $F' = f$ and $F(0) = 0$. Thus the integral becomes
\begin{align*}
&\quad \int_0^1 \left(\frac {f(x+h) - f(x)}h - f'(x)\right)\mathrm dx \\
&= \frac 1h \int_h^{1+h} f - \frac 1h \int_0^1 f - \int_0^1 f'\\
&= \frac 1h (F(1+h) - F(h) - F(1)) - (f(1) - f(0))\\
&= \frac {F(1+h) - F(1)}h - \frac {F(h) - F(0)}h - (f(1) - f(0))\\
&\xrightarrow {h \to 0} f(1) - f(0)- (f(1) - f(0))\\
& = 0
\end{align*}
as desired.
A: Claim: given $\epsilon >0$ there exists $\delta >0$ such that $|\frac {f(x+h)-f(x)} h -f'(x)| <\epsilon$ whenever $|h| <\delta$. Once this is proved the result follows immediately.
To prove this claim use MVT. We can write $|\frac {f(x+h)-f(x)} h -f'(x)| =|f'(y)-f'(x)|$ for some $y$ between $x$ and $x+h$. Note that if $|h| <\delta$ with $\delta <\frac 1 2$ then $0\leq x \leq 1$ implies $-\delta \leq x+h \leq 1+\delta$, so $-\frac 1 2 \leq x \leq \frac 3 2$. Also $[-\frac 1 2,\frac 3 2] \subset (-2,2)$. Now can you complete the proof using the fact that $f'$ is uniformly continuous on $[-\frac 1 2,\frac 3 2]$?
