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anonymous

  • 5 years ago

another question on limits

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  1. anonymous
    • 5 years ago
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    \[\lim_{x \rightarrow \infty} (\int\limits_{0}^{x}e ^{x}dx)^{2}/\int\limits_{0}^{x}e ^{2x ^{2}}dx\]

  2. apples
    • 5 years ago
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    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
    • 5 years ago
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    If you could show that the limit of the denominator goes to infinity, you could apply L'Hopital's.

  4. anonymous
    • 5 years ago
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    how about using l'hospital's rule?

  5. anonymous
    • 5 years ago
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    I think l'hospital's rule would solve it.

  6. anonymous
    • 5 years ago
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    well whats the integration of the denominator?

  7. apples
    • 5 years ago
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    There is no elementary solution to it.

  8. anonymous
    • 5 years ago
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    You can just apply the fundamental theorem of calculus to find the derivative of numerator and denominator.

  9. apples
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    That works. Then after application of FToC and L'Hopital's this should be pretty easy to solve.

  12. anonymous
    • 5 years ago
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    can any one of you solve it.. i have no idea whatsoever...lol

  13. anonymous
    • 5 years ago
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    Well, I got the answer to be 0. Is it right at first?

  14. anonymous
    • 5 years ago
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    right on buddy..=)

  15. anonymous
    • 5 years ago
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    Good.

  16. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    how is that so?

  18. anonymous
    • 5 years ago
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    Just take it for now :)

  19. anonymous
    • 5 years ago
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    alright go on..=)

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

  21. anonymous
    • 5 years ago
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    That's the derivative of the numerator.

  22. anonymous
    • 5 years ago
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    Are you following so far?

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

  24. anonymous
    • 5 years ago
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    doesn't the anti derivative come out to be \[\left(e^x - 1\right)^2\]

  25. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    oh ok i get it..=)

  27. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    \[{d \over dx}({\int\limits\limits_{0}^{x}e^t d. t })^2=2\int\limits\limits_{0}^{x}e^t d t. e^x\]

  29. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    I gotta go now. I may catch up with you later. Bye!!

  35. anonymous
    • 5 years ago
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    oh ya...how did i miss that! thanx a lot bro.. =)=)

  36. anonymous
    • 5 years ago
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    cyaaaa..

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