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This question is in the section about definite integrals and the task is to calculate the limit. My first idea was division-by-zero but I am very unsure about this. What is the goal here? I then thought that should I investigate things by different limits?

I have simplified this question but similar questions on the page 548 6* here.

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    Sure you've seen [this theorem](http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part) if you've gotten that far? The limit is the derivative of $\int_0^x \exp t^2 dt$ at $x=0$.2012-02-07

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You may re-write what you have as

$ \lim_{x\to 0}\frac{1}{x-0}\int_0^x e^{t^2}\,dt. $

If you haven't seen it before,

$ \frac{1}{x-0}\int_0^x e^{t^2}\,dt $

is the average value of the function $e^{t^2}$ over the interval $[0,x]$. Now, imagine that F'(t)=e^{t^2}. Then by the Fundamental Theorem of Calculus we have

$ \int_0^x e^{t^2}\,dt=F(x)-F(0). $

Thus, your limit becomes

$ \lim_{x\to 0}\frac{F(x)-F(0)}{x-0}. $

This is just the definition of the derivative of $F$ evaluated at $x=0$. But, we know what the derivative of $F(x)$ is, namely $e^{x^2}$.

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    @hhh: Do you mean that to be $F($x$^2)$? In that instance, $y$ou've got some chain rule stuff going on.2012-02-08
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HINT: Let $f(x)=\int_0^x e^{t^2}dt\;.$

  1. What is $\lim\limits_{x\to 0}f(x)$?
  2. What is f\,'(x)?
  3. L’Hospital’s rule.
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    @anon: Because it works even if you don’t see that you’re dealing with a difference quotient. It’s true that you don’t need it, but by that point in the course it should be a fairly basic part of your toolkit, and since it obviously works, you could easily use it without even noticing that it’s not necessary $-$ as in fact I did.2012-02-07
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i think that our integral should be understood as the mean value of the exponential on the interval $ (0,x)$ since $ x \rightarrow 0 $ the mean value on the interval $ (0,0) $ is just $ exp(0)=1$ to $1$ is the answer

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    +1 good observation, indeed that is the straightforward interpretation.2019-02-05