anonymous 5 years ago another question on limits

1. anonymous

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

2. 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.

3. apples

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

4. anonymous

5. anonymous

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

6. anonymous

well whats the integration of the denominator?

7. apples

There is no elementary solution to it.

8. anonymous

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

9. 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.

10. 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.

11. apples

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

12. anonymous

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

13. anonymous

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

14. anonymous

right on buddy..=)

15. anonymous

Good.

16. 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: ${d \over dx}{\int\limits_{0}^{x}f{(t)}d t}=f(x)$

17. anonymous

how is that so?

18. anonymous

Just take it for now :)

19. anonymous

alright go on..=)

20. 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.

21. anonymous

That's the derivative of the numerator.

22. anonymous

Are you following so far?

23. anonymous

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

24. anonymous

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

25. 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.

26. anonymous

oh ok i get it..=)

27. 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

28. anonymous

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

29. 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.

30. 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 :)

31. 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.

32. anonymous

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

33. anonymous

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

34. anonymous

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

35. anonymous

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

36. anonymous

cyaaaa..