# Integral inside a limit?

Problem gives me a review of FTC2, saying $$\frac{d}{dx}(\int_{a}^{x}f(t)\,dt) = f(x)$$ then tells me:

For $$f(t)$$ continuous, calculate the limit.

$$\lim_{h\to 0}1/h\int_{h}^{2h}f(t)\,dt$$

I started by putting it in the format of $$\lim_{h\to 0}\frac{F(2h)-F(h)}{h}$$, but now I am kind of stuck. I'm not sure what to do next. Thank you for any help in advance!

• Subtract and add 0 in the numerator and split the fraction – imranfat Jan 20 at 1:33

Define a function $$F:x \mapsto \int_{a}^{x} f(t) \ dt$$, where $$f$$ is continuous on an interval $$[a,b]$$ and $$x \in [a,b]$$. In that case, $$f$$ is Riemann Integrable and $$F$$ is differentiable so it is continuous. Now, you correctly deduced that you have to calculate the limit:

$$\lim_{h \to 0} \frac{F(2h)-F(h)}{h}$$

Observe that this is just:

$$\lim_{h \to 0} \frac{F(2h)-F(0)+F(0)-F(h)}{h} = \lim_{h \to 0} \left(\left( 2 \cdot \frac{F(2h)-F(2 \cdot 0)}{2h} \right) - \left( \frac{F(h)-F(0)}{h} \right) \right)$$

The first limit is just $$2f(0)$$ and the second limit is just $$f(0)$$. Hence, the entire limit evaluates to $$f(0)$$ and we are done.

• Do you mean $f(0)$? – Nicholas Roberts Jan 20 at 1:39
• Ah yes, thanks for spotting the error :D – Abhi Jan 20 at 1:41

Apply l'Hopital's rule.

$$\lim_{h\to0}\frac{F(2h)-F(h)}{h}$$ $$=\lim_{h\to0}\frac{\frac d{dh}[F(2h)-F(h)]}{\frac d{dh}h}$$ $$=\lim_{h\to0}\frac{\frac d{dh}F(2h)-\frac d{dh}F(h)}1$$ $$=\lim_{h\to0}\frac d{dh}[2h]F'(2h)-F'(h)$$ $$=\lim_{h\to0}2F'(2h)-F'(h)$$

Now recall $$F'(x)=f(x)$$. Replace $$h$$ with $$0$$ and simplify.