# Calculating limit of integral using L'Hospital's rule

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?

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

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

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 : $$\lim\limits_{h\rightarrow 0}\frac{f(3+h)-f(3)}{h}=\lim\limits_{h\rightarrow 0}\frac{f'(3+h)}{1}=f'(3)=e^9$$

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

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