# Define $\lim_{x\rightarrow 0} \frac{1}{x}\int_{0}^{x} e^{t^{2}} dt$, what is the purpose of this question?

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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The goal is to see if you have understood the material covered till the fundamental theorem of calculus. –  Aryabhata Feb 7 '12 at 22:15
Which book did you get this from? –  Aryabhata Feb 7 '12 at 22:22
Sure you've seen this theorem if you've gotten that far? The limit is the derivative of $\int_0^x \exp t^2 dt$ at $x=0$. –  anon Feb 7 '12 at 22:29

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

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

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

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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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 7 '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 7 '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. –  Joe Johnson 126 Feb 8 '12 at 14:17

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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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 7 '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 7 '12 at 23:03
@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 7 '12 at 23:07
In my opinion, that would defeat the point of the exercise. –  anon Feb 7 '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 7 '12 at 23:13

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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Did you ask the author of the question about his intended method of solution/interpretation? –  The Chaz 2.0 Apr 22 '12 at 13:33
+1 good observation, indeed that is the straightforward interpretation. –  hhh Apr 22 '12 at 14:06