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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!

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    $\begingroup$ Subtract and add 0 in the numerator and split the fraction $\endgroup$ – imranfat Jan 20 at 1:33
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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.

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

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