# Why does $\lim_{h \to 0} \frac{1}{h} \int_{b}^{b+h} f = f(b)$?

I've been reading a book lately and this has come up several times. If $f$ is continuous, then $$\lim_{h \to 0} \frac{1}{h} \int_{b}^{b+h} f = f(b).$$ Now, I know I can use the fundamental theorem of calculus to show this, and I just wanted to check that I'm not missing a much simpler argument. I ask because this is always said to 'follow from continuity' and the fundamental theorem of calculus is never explicitly mentioned.

• I think the Fundamental theorem of Calculus would be a good resort indeed – imranfat Apr 14 '16 at 13:44
• In fact the result you have mentioned above is called First Fundamental Theorem of Calculus namely if $f$ is Riemann integrable on $[a, b]$ and $$F(x) = \int_{a}^{x}f(t)\,dt$$ then $$F'(c) = \lim_{h \to 0}\frac{1}{h}\int_{c}^{c + h}f(t)\,dt = f(c)$$ for all points $c \in [a, b]$ at which $f$ is continuous. – Paramanand Singh Apr 17 '16 at 15:03

## 1 Answer

Let $\epsilon>0$. Choose $\delta>0$ such that etc. If $0<h<\delta$ then

$$\left|\frac1h\int_b^{b+h}f(t)\,dt-f(b)\right| =\left|\frac1h\int_b^{b+h}(f(t)-f(b))\,dt\right| \le \frac1h\int_b^{b+h}|f(t)-f(b)|\,dt\le\frac1h\int_b^{b+h}\epsilon\,dt=\epsilon.$$