another question on limits

- anonymous

another question on limits

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- anonymous

\[\lim_{x \rightarrow \infty} (\int\limits_{0}^{x}e ^{x}dx)^{2}/\int\limits_{0}^{x}e ^{2x ^{2}}dx\]

- apples

The numerator easily simplifies to \[ \left(e^x - 1\right)^2 \] but the lower integral has no elementary antiderivative, so I'm not too sure.

- apples

If you could show that the limit of the denominator goes to infinity, you could apply L'Hopital's.

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## More answers

- anonymous

how about using l'hospital's rule?

- anonymous

I think l'hospital's rule would solve it.

- anonymous

well whats the integration of the denominator?

- apples

There is no elementary solution to it.

- anonymous

You can just apply the fundamental theorem of calculus to find the derivative of numerator and denominator.

- apples

Yes, but L'Hopital's only works if you know you have an indeterminate form. We don't know what the limit of the denominator is.

- anonymous

I think by looking at a graph for e^(x^2), we can see that the area under this graph goes to infinity as x goes to infinity.

- apples

That works. Then after application of FToC and L'Hopital's this should be pretty easy to solve.

- anonymous

can any one of you solve it.. i have no idea whatsoever...lol

- anonymous

Well, I got the answer to be 0. Is it right at first?

- anonymous

right on buddy..=)

- anonymous

Good.

- anonymous

We're going to apply L'Hopital's rule, but you have first to be aware of the Fundamental theorem of calculus, that states:
\[{d \over dx}{\int\limits_{0}^{x}f{(t)}d t}=f(x)\]

- anonymous

how is that so?

- anonymous

Just take it for now :)

- anonymous

alright go on..=)

- anonymous

\[{d \over dx}({\int\limits_{0}^{x}e^t d. t })^2=2\int\limits_{0}^{x}e^t d t. e^x=2e^x(e^x-1)\]
I used chain rule here as well.

- anonymous

That's the derivative of the numerator.

- anonymous

Are you following so far?

- anonymous

The derivative of the denominator is just:
\[{d \over dx}\int\limits_{0}^{x}e^{2t^2} d.t=e^{2x^2}\]

- anonymous

doesn't the anti derivative come out to be \[\left(e^x - 1\right)^2\]

- anonymous

Well what you wrote above is the value of the integral, but we're looking for the derivative of the square of the integral. That's equal to, by chain rule, 2*the integral*its derivative.

- anonymous

oh ok i get it..=)

- anonymous

can you explain this a little.. {d \over dx}({\int\limits_{0}^{x}e^t d. t })^2=2\int\limits_{0}^{x}e^t d t. e^x

- anonymous

\[{d \over dx}({\int\limits\limits_{0}^{x}e^t d. t })^2=2\int\limits\limits_{0}^{x}e^t d t. e^x\]

- anonymous

Good. Now we have found both derivatives of the numerator and the denominator. So, we have:
\[{2e^{2x}-2e^x \over e^{2x^2}}={2e^{2x}(1-{1 \over e^x}) \over e^{2x}(e^{2x^2-2x})}={2-{2 \over e^x} \over e^{2x^2-2x}}\]
The top goes to 2 as x goes to infinity, and the bottom goes to infinity as x goes to infinity. Hence the limit goes to 0.

- anonymous

The key to find what you asked about is applying both chain rule and FToC at the same time. So for chain rule assume the integral is u, then: d/du(u^2)=2u.u'. u is the value of the integral which is (e^x-1). u' is the derivative of the integral which, by FToC, e^x. ans 2 is just 2 :)

- anonymous

To understand the FToC, look at the differentiation, for now, as the inverse operation of integration. So they "undo" each other. I hope that helps.

- anonymous

hmm..got it.. but i didn't understand how you simplified the denominator here \[{2e^{2x}-2e^x \over e^{2x^2}}={2e^{2x}(1-{1 \over e^x}) \over e^{2x}(e^{2x^2-2x})}={2-{2 \over e^x} \over e^{2x^2-2x}}\]

- anonymous

When you take something is a common factor, you divide the rest by it. Right?
When dividing you subtract powers, that's all.

- anonymous

I gotta go now. I may catch up with you later. Bye!!

- anonymous

oh ya...how did i miss that!
thanx a lot bro.. =)=)

- anonymous

cyaaaa..

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