# How to handle $\lim_{h \rightarrow 0} \frac{1}{h} \int_{0}^h f(s) ds$

In my lecture notes I found the following identity: $$\lim_{h \rightarrow 0} \frac{1}{h} \int_{0}^h f(s) ds = f(0)$$ without any explanation on what's going on here. I think in the lecture we said that $f$ should be a $C^1$-function but I'm not sure, whether one needs this assumption or not. My question now is: Why does this identity hold? I was thinking of using the Fundamental Theorem of Calculus - but how do I know that there is a function $F$ s.t. $F'(x) = f(x)$?

Thanks for any help!

• Many ways to show this. One route is to use L'Hopitals rule + the fundamental theorem of calculus. Dec 23, 2015 at 12:45
• If f is integrable, then the integral is F(s)-F(0). Take a look of the definition of derivative. Dec 23, 2015 at 12:45
• Yes, I know that it is clear, that the identity holds if there is a function $F$ as described above. But does such an $F$ exist? Dec 23, 2015 at 12:49
• The fundamental theorem needs f to be continuous. Dec 23, 2015 at 12:57
• The required condition is that f has to be conituous at 0 for such a function F to exist...... Dec 23, 2015 at 12:57

It is possible to prove this limit whithout using derivatives, but only using the continuity of $f$. We want to show that $$\left|\frac{1}{h}\int_0^h f(s) ds-f(0)\right|<\epsilon$$ when $h<\delta$ for some $\delta>0$. Now $$\left|\frac{1}{h}\int_0^h f(s) ds-f(0)\right|=\left|\frac{1}{h}\int_0^h [f(s)-f(0)] ds\right|\le \frac{1}{h}\int_0^h |f(s)-f(0)| ds.$$ I used the fact that $\frac{1}{h}\int_0^h f(0)ds=f(0)$. Since $f$ is continuous, for every $\epsilon$, there exists $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ when $|x-y|<\delta$. Now if we take $h<\delta$, then $|f(s)-f(0)|<\epsilon$ when $s \in (0,h)$, hence $$\frac{1}{h}\int_0^h |f(s)-f(0)| ds\le \frac{1}{h}\int_0^h \epsilon ds=\epsilon.$$