Verify Functional Limit of $\lim_{h \to 0} \frac{\int_{1-h}^{1+h} f(x) \,dx}{h}$ I want to find $$\lim_{h \to 0} \dfrac{\displaystyle\int_{1-h}^{1+h} f(x) \,dx}{h}$$. Where $f$ is a continuous function defined on $\mathbb{R}$
What I have so far:
Since $$\lim_{h \to 0} \int_{1-h}^{1+h} f(x) \,dx = 0 \text{ and } \lim_{h \to 0} h = 0$$ we can use L'Hopital's rule.
$\dfrac{d}{dh}h =1$ and $\frac{d}{dh} \displaystyle\int_{1-h}^{1+h} f(x) \,dx = f(1+h)+f(1-h)$. (Is this part right?).
Then from that we apply L'Hopital's rule:
$$\lim_{h \to 0} \frac{\displaystyle \int_{1-h}^{1+h} f(x) \,dx}{h} = \lim_{h \to 0} f(1+h)+f(1-h) = f(1) +f(1)=2f(1)$$
Does this look right? Thanks!
 A: By mean value theorem, for all $h>0$, there is always a number $x_h \in (1-h,1+h)$ such that:
$\int^{1+h}_{1-h} f(x)dx= 2h f(x_h)$
And as $h$ converges to $0$, clearly, $x_h$ converges to one. Hence by continuity, the desired limit is $2f(1)$.
A: Checking the part you aren't sure: $$\frac{d}{dh} \int_{1-h}^{1+h} f(x) \,dx = \frac{d}{dh}\left(\int_{0}^{1+h} f(x) \ dx - \int_{0}^{1-h} f(x) \ dx\right) = \frac{d}{dh}\int_{0}^{1+h} f(x) \ dx  - \frac{d}{dh} \int_{0}^{1-h} f(x) \ dx = f(1+h) - (-f(1-h)).$$
Obtaining thus the result that you want. Then, using the L'hopital rule that you're mentioning, the result follows. Your answer is correct.
A: Alternative solution
Let $g(x) = \displaystyle\int\limits_{0}^{x} f(t) \ dt$, then $\displaystyle\int\limits_{1-h}^{1+h} f(x) \ dx = g(1+h)-g(1-h)$. Therefore, $\lim\limits_{h\rightarrow 0}\frac{\int_{1-h}^{1+h}f(x)\ dx}{h}=\lim\limits_{h\rightarrow 0}\frac{g(1+h)-g(1-h)}{h}=\lim\limits_{h\rightarrow 0}\left(\frac{g(1+h)-g(1)}{h}+\frac{g(1)-g(1-h)}{h}\right)=2g'(1)=2f(1).$
I think Paresseux has given the most beautiful solution, though.
A: Alternative solution 2
Since $f$ is continuous, for all $\epsilon>0$ there exists $\delta>0$ such that, if $1-\delta<x<1+\delta$, then $f(1)-\frac{\epsilon}{2}<f(x)<f(1)+\frac{\epsilon}{2}$. From this follows that $2h(f(1)-\frac{\epsilon}{2}) < {\displaystyle\int\limits_{1-h}^{1+h}f(x) \ dx} < 2h(f(1)+\frac{\epsilon}{2})$ for all $0<h<\delta$, which is equivalent to $2f(1)-\epsilon < \frac{\int_{1-h}^{1+h}f(x) \ dx}{h} < 2f(1)+\epsilon$ for all $0<h<\delta$ (it can easily be checked that the same holds when $-\delta<h<0$). Thus the limit follows directly from the definition.
