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Math Help - Please help me with antiderivative/l'hopitale problems

  1. #1
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    Please help me with antiderivative/l'hopitale problems

    Id highly appreciate it if anyone could help me with any of the following
    problems!

    What is the antiderivative of (cosx)^4 (cos(x^4))??




    Find lim as x->infinity [(2x-3)/(2x+5)]^2x+1




    Find lim as x->infinity [e^(x^2) - e^ [(x^2) + xsinx)] ]




    Find f ' (x) and f '' (x) if f(x)= e^(-1/sin^2x) ,where x is between negative pi/2 and positive pi/2
    f(x)= 0 , where x = 0

    Thanks in advance!
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by shmayer View Post
    Id highly appreciate it if anyone could help me with any of the following
    problems!

    What is the antiderivative of (cosx)^4 (cos(x^4))??




    Find lim as x->infinity [(2x-3)/(2x+5)]^2x+1




    Find lim as x->infinity [e^(x^2) - e^ [(x^2) + xsinx)] ]




    Find f ' (x) and f '' (x) if f(x)= e^(-1/sin^2x) ,where x is between negative pi/2 and positive pi/2
    f(x)= 0 , where x = 0

    Thanks in advance!
    \int\cos^4(x)\cos(x^4)\,dx has no closed from solution.

    --------------

    There might be a slick way to do this, but here's a method that will work, though it's a bit tedious and may cause retinal bleeding.

    \lim_{x\to\infty}\left(\frac{2x-3}{2x+5}\right)^{2x+1}=\lim_{x\to\infty}\exp\left[(2x+1)\ln\left(\frac{2x-3}{2x+5}\right)\right]= \exp\left[\lim_{x\to\infty}\frac{\ln\left(\frac{2x-3}{2x+5}\right)}{(2x+1)^{-1}}\right]

    Use L'Hopital's rule on that limit.

    --------------

    \lim_{x\to\infty}e^{x^2}-e^{x^2+x\sin x}

    This limit is undefined, because e^{x^2}-e^{x^2+x\sin x}=e^{x^2}\left(1-e^{x\sin x}\right). Since \lim_{x\to\infty}x\sin x DNE, the whole limit DNE.

    --------------

    f(x)=\exp\left(\frac{-1}{\sin^2x}\right)

    For these derivatives, just use the chain rule.
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  3. #3
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    hey, thanks for the reply..for the one question you said it DNE since limx->infinity (xsinx) DNE...but couldnt i use Lhopitales rule before you factored out e^x ?
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  4. #4
    Super Member redsoxfan325's Avatar
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    L'Hopital's rule requires \frac{0}{0} or \frac{\infty}{\infty}. We didn't have either of those.
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  5. #5
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    Cant we use lhopitales rule for infinity - infinity? that is indeterminate form right?
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by shmayer View Post
    Cant we use lhopitales rule for infinity - infinity? that is indeterminate form right?
    No, it doesn't work for that form. For example, consider:

    \lim_{x\to\infty}(x-\ln x)

    This is \infty-\infty, and the whole limit equals \infty.

    But if you try to use L'Hopital's rule, you get \lim_{x\to\infty}\left(1-\frac{1}{x}\right)=1, which is wrong.
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  7. #7
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    Ah i see thanks! Also, why does the first question not have a solution?
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  8. #8
    Super Member redsoxfan325's Avatar
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    There's no solution that can be expressed in terms of elementary functions whose derivative is \cos(x^4)\cos^4(x).
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  9. #9
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    Quote Originally Posted by redsoxfan325 View Post
    There's no solution that can be expressed in terms of elementary functions whose derivative is \cos(x^4)\cos^4(x).
    Why is this the case?
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  10. #10
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    also, how would i differentiate f and find the domain of f(x)= lnlnlnx ?

    Thanks a TON!
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  11. #11
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by shmayer View Post
    Why is this the case?
    Because none of the methods used to integrate (u-sub, by parts, etc.) yield anything integrable. I know that's a pretty lousy answer; maybe someone else on MHF can explain it better.

    Quote Originally Posted by shmayer View Post
    also, how would i differentiate f and find the domain of f(x)= lnlnlnx ?

    Thanks a TON!
    For differentiation, you have to use the chain rule.

    For the domain, \ln(u) is defined for u>0. In this case, u=\ln(\ln(x)), so we need \ln(\ln(x))>0. This is the case when \ln(x)>1. (Do you see why?) And \ln(x)>1 when x>e. So the domain of \ln(\ln(\ln(x))) is (e,\infty).
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  12. #12
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    Thanks!
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