Limit in integral: $\lim\limits_{\epsilon \rightarrow 0}\frac1{\epsilon}\int_{t}^{t+\epsilon}f(s)ds=f(t)$? Let $f$ be a smooth function  : 
can someone tell me why we have :
$\lim\limits_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int_{t}^{t+\epsilon}f(s)ds=f(t)$ 
thank you very much !

Let $F(x) = \int_0^x f$. Then $F$ is smooth since $f$ is, and $F^\prime = f$ by the Fundamental Theorem of Calculus; but we also have $$F^\prime(t) = \lim_{\varepsilon\to0}\frac{F(t+\varepsilon) - F(t)}{\varepsilon} = \lim_{\varepsilon\to0}\frac{\int_t^{t+\varepsilon} f}{\varepsilon}.$$
