Evaluation of integral Let $ f \in C[-1,1]$. Evaluate: 
$$\lim_{h \to  0} \frac{ \int^{h}_{-h} f(t) \, dt }{h} $$
My attempt:- Using L' Hospital's rule, 
$$\lim_{h \to  0} \frac{\frac{d}{dh} \int^{2h}_{0} f(t) \, dt }{1}  =  \lim_{h \to 0}   f(2h)  = f(0)$$
Am I correct?
 A: It seems that using l'Hopital for this is circular (or nearly circular).  Why not use more elementary means?  
Let $\epsilon > 0$.  Since $f$ is continuous at zero, there is $\delta > 0$ so that if $|t-0|<\delta$, then $|f(t)-f(0)|<\epsilon$.  Now et $0<h<\delta$.  Then every $t \in [-h,h]$ satisfies $|t-0|<\delta$, so
$$
(f(0)-\epsilon)\cdot 2h \le 
\int_{-h}^h f(t)\;dt \le (f(0)+\epsilon)\cdot 2h
\\
2(f(0)-\epsilon) \le 
\frac{1}{h}\int_{-h}^h f(t)\;dt \le 2(f(0)+\epsilon)
$$
So we conclude
$$
\lim_{h \to 0^+} \frac{1}{h}\int_{-h}^h f(t)\;dt = 2f(0) .
$$
A: Another approach would be to use the Mean Value Theorem for Integrals:
Since $\int_{-h}^{h} f(t)\, dt= 2hf(c_{h})$ for some $c_{h}$ in $(-h,h)\;\;\;$ (or $(h,-h)$, if $h<0$), 
$$\lim_{h \to  0} \frac{ \int^{h}_{-h} f(t) \, dt }{h}=\lim_{h\to 0} \frac{2hf(c_{h})}{h}=\lim_{h\to 0}2f(c_{h})=2f(0).$$
$------------------------------------------\\$
Here is a similar argument which does not use the MVT for Integrals:
We can assume that $h>0$, since the function is an even function of h; so
let $M_h$ and $m_h$ be the maximum and minimum values of f on $[-h,h]$.
Then $2hm_h\le \int_{-h}^{h} f(t)\;dt\le 2hM_h\;\;$ and
$\displaystyle\;\;2m_h\le \frac{ \int^{h}_{-h} f(t) \, dt }{h}\le 2M_h $ for $h>0$.
Since $\displaystyle \lim_{h\to 0}2m_h=2f(0)$ and $\displaystyle \lim_{h\to 0}2M_h=2f(0)\;\;$ since f is continuous at 0,
$\displaystyle\;\;\;\lim_{h \to  0} \frac{ \int^{h}_{-h} f(t) \, dt }{h}=2f(0)$ by the Squeeze Theorem.
A: Assuming $\;f\;$ is continuous and thus has a primitive function $\;F\;$ on $\;[-1,1]\;$ , we'd get applying l'Hospital's rule:
$$\lim_{h\to 0}\frac1h\int\limits_{-h}^hf(t)dt=\lim_{h\to 0}\frac{F(h)-F(-h)}h\stackrel{\text{l'H}}=\lim_{h\to 0}(F'(h)+F'(-h))=$$
$$=\lim_{h\to 0}(f(h)+f(-h))=2f(0)$$
and we do not get what you got in this particular case...
A: Hint: In your attempt, you just changed the limits of integration without justification, but tree to see whether you can somehow rewrite your integral into a sum or difference of integrals whose lower limits are constant.  That way you can easily apply the fundamental theorem of calculus.  And don't forget to use the chain rule if the need arises.
A: L'Hopital's Rule is valid here, and while there are other ways to do the problem, L'Hopital's Rule works fine.  Hint: you have the right idea, but you changed the limits of integration for no reason, and failed to apply the Chain Rule.  You need to use the Fundamental Theorem of Calculus and the Chain Rule.  You get:
$$\frac{d}{dh} \int_{-h}^h f(t)\,dt= f(h)\frac{d}{dh}(h)-f(-h)\frac{d}{dh}(-h)=f(h)+f(-h).$$
