Let be $f$ a continuous function. Determine the limit $\lim\limits_{h \to 0} \frac{1}{h} \int_{a-h}^{a+h} f(x)\,dx$ $\lim\limits_{h \to 0} \frac{1}{h} \int_{a-h}^{a+h} f(x)\,dx$
I think that this kind of limit should I probably calculate with some kind of epsilon-delta definition. 
And using the limits:
$\lim\limits_{h \to 0^+} \frac{1}{h}=\infty $
$\lim\limits_{h \to 0^-} \frac{1}{h}=-\infty $
I appreciate any helps.
 A: Note $f$ has an antiderivative, $F$. Thus $\int_{a-h}^{a+h} f(t) \ dt = F(a+h) - F(a-h)$ and so we're evaluating 
$$\lim_{h \to 0} \frac{F(a+h) - F(a-h)}{h} = 2\lim_{h \to 0} \frac{F(a+h) - F(a-h)}{2h} = 2F'(a) = 2f(a)$$
Note that the second-to-last equality follows from the fact that if the derivative of a function exists, its symmetric derivative exists and is equal to it. 
A: Using L'Hospital's rule we get
\begin{align*}
\lim_{h \to 0}\frac{\int_{a-h}^{a+h}f(x) \,dx}{h} &= 
\lim_{h \to 0}\frac{f(a+h)+f(a-h)}{1} \\
&= 2f(a).
\end{align*}
A: I will first prove for $h\rightarrow 0^{+}$, for $h\rightarrow 0^{-}$ is treated simialrly. As $f$ is continuous at $x=a$, given $\epsilon>0$, one may find some $\delta>0$ such that $|f(x)-f(a)|<\epsilon$ for every $x$ with $|x-a|<\delta$. For all $h\in(0,\delta)$, we have 
$\left|\dfrac{1}{h}\displaystyle\int_{a-h}^{a+h}f(x)dx-2f(a)\right|\\
=\left|\dfrac{1}{h}\displaystyle\int_{a-h}^{a+h}f(x)dx-\dfrac{1}{h}\displaystyle\int_{a-h}^{a+h}f(a)dx\right|\\
\leq\dfrac{1}{h}\displaystyle\int_{a-h}^{a+h}|f(x)-f(a)|dx\\
\leq\dfrac{1}{h}\displaystyle\int_{a-h}^{a+h}\epsilon dx\\
=2\epsilon$.
So the limit is $2f(a)$.
