Question

another question on limits

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

how about using l'hospital's rule?

Answer 5

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

Answer 6

well whats the integration of the denominator?

Answer 7

There is no elementary solution to it.

Answer 8

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

Answer 9

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.

Answer 10

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.

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

right on buddy..=)

Answer 15

Good.

Answer 16

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

Answer 17

how is that so?

Answer 18

Just take it for now :)

Answer 19

alright go on..=)

Answer 20

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

Answer 21

That's the derivative of the numerator.

Answer 22

Are you following so far?

Answer 23

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

Answer 24

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

Answer 25

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.

Answer 26

oh ok i get it..=)

Answer 27

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

Answer 28

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

Answer 29

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.

Answer 30

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 :)

Answer 31

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.

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

cyaaaa..