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Calculate: $$\lim_{h\rightarrow 0} \frac{1}{h}\int_3^{3+h}e^{t^2}dt$$ I was thinking of using L'Hospital's rule, as this limit is of form $\frac{0}{0}$.

So the limit above equals(after derivatives): $$\lim_{h\rightarrow 0} \frac{e^{h^2}}{1}=1$$

Is it correct?

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No. If $G(h)=\int_3^{3+h}e^{t^2}dt$, then $G'(h)=e^{(3+h)^2}.$

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The lmit is $\lim_{h \to 0}\frac {e^{(3+h)^{2}}} 1=e^{9}$.

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No this is false, let $f(h)=\int_0^h e^{t^2}dt$, then $\lim\limits_{h\rightarrow 0}\frac{f(3+h)-f(3)}{h}=f'(3)=e^9$. You can also use L'Hopital's rule : $$ \lim\limits_{h\rightarrow 0}\frac{f(3+h)-f(3)}{h}=\lim\limits_{h\rightarrow 0}\frac{f'(3+h)}{1}=f'(3)=e^9 $$

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You made a mistake in the limit it should be $e^{(3+h)^2}$

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From L'Hospital's rule, we have

$\lim _{h\rightarrow 0}\left(\frac{\int _3^{3+h}\left(e^{t^2}\right)dt}{h}\right)=\frac{\frac{d}{dh}\left(\int _3^{3+h}\left(e^{t^2}\right)dt\right)}{\frac{d}{dh}\left(h\right)}$

From the Newton-Leibniz' Theorem--

$\frac{d}{dh}\left(\int _3^{3+h}\left(e^{t^2}\right)dt\right)=e^{\left(3+h\right)^2}\left(\frac{d}{dh}\left(3+h\right)\right)=e^{\left(3+h\right)^2}$

Hope it is clear now :~) Still if any query feel free to ask

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