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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.

hhh
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3 Answers3

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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 $e^{x^2}$.

J126
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  • What about if the other limit is of different form such as $x^{7}$, $ln(x)$ --? Then, it is not exactly the theorem (or the average -analogy). I think I need to do some adjusting or investigation of the limit then somehow? – hhh Feb 07 '12 at 22:53
  • For example, this is not for the def. of derivative so do I need to somehow adjust it? $$\lim_{x\rightarrow 3} \frac{x^{2}-F(9)}{x-3}$$ – hhh Feb 07 '12 at 22:59
  • @hhh: Do you mean that to be $F(x^2)$? In that instance, you've got some chain rule stuff going on. – J126 Feb 08 '12 at 14:17
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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.
Brian M. Scott
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    Perhaps the downvoter would care to explain? The suggested argument is in fact both correct and easy, and the answer has the virtue of not completely doing the homework problem for the OP. – Brian M. Scott Feb 07 '12 at 22:59
  • I don't see the point of #3 but I don't get the downvote either. This is basically what I would have said. – anon Feb 07 '12 at 23:03
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    @anon: The point of (3) is that you don’t have to recognize this as the limit of a difference quotient: you can also see it simply as a $0/0$ indeterminate form. – Brian M. Scott Feb 07 '12 at 23:07
  • In my opinion, that would defeat the point of the exercise. – anon Feb 07 '12 at 23:10
  • @anon: I take a different view of the point of the exercise: I think that the point is the fundamental theorem, as in my point (2). – Brian M. Scott Feb 07 '12 at 23:13
  • @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. – Brian M. Scott Feb 07 '12 at 23:21
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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

Jose Garcia
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