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Math Help - Limit Marathon

  1. #61
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    Try this one: \lim_{\alpha\to\infty}\frac1{\alpha^2}\int_0^\alph  a\ln(1+e^{x})\,dx.
    Your going to hate me for ruining your beautiful limit, but

    \lim_{\alpha\to\infty}\frac{\int_0^{\alpha}\ln(1+e  ^x)}{\alpha^2}=\lim_{\alpha\to\infty}\frac{\ln(1+e  ^{\alpha})}{2\alpha}

    now \ln(1+e^{\alpha})\sim\ln(e^{\alpha})=\alpha

    thus the limit is \lim_{\alpha\to\infty}\frac{1}{\alpha^2}\int_0^{\a  lpha}\ln(1+e^x)dx=\frac{1}{2}

    by L'hopitals
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  2. #62
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    That's the correct answer, but find another solution without that rule.
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  3. #63
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    That's the correct answer, but find another solution without that rule.
    Am I on the right track (whited out)

    convert the integral to a Riemann sum?
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  4. #64
    Super Member PaulRS's Avatar
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    \alpha\cdot{\int_0^1\ln(1+e^{\alpha\cdot{x}})dx}=\  int_0^{\alpha}\ln(1+e^x)dx

    Thus: \frac{\int_0^{\alpha}\ln(1+e^x)dx}{\alpha^2}=\frac  {\int_0^1\ln(1+e^{\alpha\cdot{x}})dx}{\alpha}

    So our limit is: \lim_{\alpha\rightarrow{+\infty}}\frac{\int_0^1\ln  (1+e^{\alpha\cdot{x}})dx}{\alpha}= \int_0^1\lim_{\alpha\rightarrow{+\infty}}\frac{\ln  (1+e^{\alpha\cdot{x}})}{\alpha}dx=\int_0^1xdx =\frac{1}{2}
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  5. #65
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    \alpha\cdot{\int_0^1\ln(1+e^{\alpha\cdot{x}})dx}=\  int_0^{\alpha}\ln(1+e^x)dx

    Thus: \frac{\int_0^{\alpha}\ln(1+e^x)dx}{\alpha^2}=\frac  {\int_0^1\ln(1+e^{\alpha\cdot{x}})dx}{\alpha}

    So our limit is: \lim_{\alpha\rightarrow{+\infty}}\frac{\int_0^1\ln  (1+e^{\alpha\cdot{x}})dx}{\alpha}= \int_0^1\lim_{\alpha\rightarrow{+\infty}}\frac{\ln  (1+e^{\alpha\cdot{x}})}{\alpha}dx=\int_0^1xdx =\frac{1}{2}
    Is there some theorem that says your first change with the integral works, or did you just find it by inpsection?
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  6. #66
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    Okay, here's another solution:

    \int_{0}^{\alpha }{\ln \left( 1+e^{x} \right)\,dx}=\frac{1}{2}\alpha ^{2}+\int_{0}^{\alpha }{\ln \left( 1+e^{-x} \right)\,dx}.

    Since \ln \left( 1+e^{-x} \right) is comparable to e^{-x} for large x, the second integral is bounded for all \alpha, then as \alpha\to\infty we have \int_{0}^{\alpha }{\ln \left( 1+e^{x} \right)\,dx}=\frac{1}{2}\alpha ^{2}+\mathcal O(1)\implies \frac{1}{\alpha ^{2}}\int_{0}^{\alpha }{\ln \left( 1+e^{x} \right)\,dx}=\frac{1}{2}+\mathcal O\left( \alpha^{-2} \right), and the limit is \frac12.

    Next problem: find \lim_{n\to\infty}\int_0^\pi\bigg|\bigg(x+\frac\pi n\bigg)\sin nx\bigg|\,dx,\,n\in\mathbb N.
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  7. #67
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    Quote Originally Posted by Mathstud28 View Post
    Try this one without L'hopitals

    \lim_{\xi\to{0}}\frac{\ln(\sin(a\xi))}{\ln(\sin(b\  xi))}
    a little trick to remember:

    \lim_{\xi\to0} \frac{\ln(\sin(a\xi))}{\ln(\sin(b\xi))}=1+\lim_{\x  i\to0} \frac{\ln(\sin(a\xi))-\ln(\sin(b\xi))}{\ln(\sin(b\xi))}=1+\lim_{\xi\to0}  \frac{\ln \left(\frac{\sin(a\xi)}{\sin(b\xi)}\right)}{\ln(\s  in(b\xi))}

    =1+\frac{\ln(\frac{a}{b})}{-\infty}=1. \ \ \ \square
    .................................................. ..................................................

    note that in order to make sure everything is defined, we must either have

    a>0, \ b>0, and \xi \rightarrow 0+, or a<0, \ b < 0, and \xi \rightarrow 0-.
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  8. #68
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    a little trick to remember:

    \lim_{\xi\to0} \frac{\ln(\sin(a\xi))}{\ln(\sin(b\xi))}=1+\lim_{\x  i\to0} \frac{\ln(\sin(a\xi))-\ln(\sin(b\xi))}{\ln(\sin(b\xi))}=1+\lim_{\xi\to0}  \frac{\ln \left(\frac{\sin(a\xi)}{\sin(b\xi)}\right)}{\ln(\s  in(b\xi))}

    =1+\frac{\ln(\frac{a}{b})}{-\infty}=1. \ \ \ \square
    .................................................. ..................................................

    note that in order to make sure everything is defined, we must either have

    a>0, \ b>0, and \xi \rightarrow 0+, or a<0, \ b < 0, and \xi \rightarrow 0-.
    So your first move was to

    \text{add  }1+-1=1+-\frac{\ln(\sin(b\xi)}{\ln(\sin(b\xi))}
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  9. #69
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    right Mathstud28, just add 1 and subtract 1. it's as simple as that!

    Quote Originally Posted by Krizalid View Post
    Next problem: find \lim_{n\to\infty}\int_0^\pi\bigg|\bigg(x+\frac\pi n\bigg)\sin nx\bigg|\,dx,\,n\in\mathbb N.
    first of all the absolute value sign is needed for \sin nx only.

    clearly \int_{k\pi}^{(k+1)\pi}y|\sin y| \ dt = \int_0^{\pi}(y+k\pi)\sin y \ dy. call this (1). now, in your integral, let nx+\pi=y. then:

    I_n=\int_0^{\pi}\left(x + \frac{\pi}{n}\right)|\sin nx| \ dx=\frac{1}{n^2}\int_{\pi}^{(n+1)\pi}y|\sin y| \ dy=\frac{1}{n^2}\sum_{k=1}^n \int_{k\pi}^{(k+1)\pi}y|\sin y|\ dy. therefore by (1):

    I_n=\frac{1}{n^2}\sum_{k=1}^n \int_0^{\pi}(y+k\pi)\sin y \ dy=\frac{1}{n}\int_0^{\pi}y\sin y \ dy \ + \ \frac{n(n+1)\pi}{2n^2}\int_0^{\pi}\sin y \ dy. therefore:

    \lim_{n\to\infty}I_n=0 \ + \ \lim_{n\to\infty} \frac{n(n+1)\pi}{n^2}=\pi. \ \ \ \square
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  10. #70
    Math Engineering Student
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    That's correct.
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  11. #71
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    right Mathstud28, just add 1 and subtract 1. it's as simple as that!



    first of all the absolute value sign is needed for \sin nx only.

    clearly \int_{k\pi}^{(k+1)\pi}y|\sin y| \ dt = \int_0^{\pi}(y+k\pi)\sin y \ dy. call this (1). now, in your integral, let nx+\pi=y. then:

    I_n=\int_0^{\pi}\left(x + \frac{\pi}{n}\right)|\sin nx| \ dx=\frac{1}{n^2}\int_{\pi}^{(n+1)\pi}y|\sin y| \ dy=\frac{1}{n^2}\sum_{k=1}^n \int_{k\pi}^{(k+1)\pi}y|\sin y|\ dy. therefore by (1):

    I_n=\frac{1}{n^2}\sum_{k=1}^n \int_0^{\pi}(y+k\pi)\sin y \ dy=\frac{1}{n}\int_0^{\pi}y\sin y \ dy \ + \ \frac{n(n+1)\pi}{2n^2}\int_0^{\pi}\sin y \ dy. therefore:

    \lim_{n\to\infty}I_n=0 \ + \ \lim_{n\to\infty} \frac{n(n+1)\pi}{n^2}=\pi. \ \ \ \square
    Have you seen this problem before..or do you just have a knack for seeing embarrassingly obvious little tricks?

    Why dont you submit a limit NonCommonAlg? That would be great, for I am sure it would be fascinating...but noting your other posts..it would preferably NOT include any of the following:
    Limits involving sums
    Limits involving sums esecially those excluding odd primes
    Limits involving integrals
    Limits involving integrals that require both countour integration and a parameterazation to solve
    Limits involving infinite products
    Limits involving anything to do with Wallis's Product
    I am sure there are quite a few more
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  12. #72
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    Quote Originally Posted by Mathstud28 View Post
    Have you seen this problem before..or do you just have a knack for seeing embarrassingly obvious little tricks?
    no i haven't, which means the "or" part of your question is probably true!
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  13. #73
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    no i haven't, which means the "or" part of your question is probably true!
    So are you going to try to oblige that request...despite the inummerable restrictions I am sure will vex you?
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  14. #74
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    Quote Originally Posted by Mathstud28 View Post
    So are you going to try to oblige that request...despite the inummerable restrictions I am sure will vex you?
    i might! right now, i don't have any that survives your tough conditions! haha
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  15. #75
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    i might! right now, i don't have any that survives your tough conditions! haha
    Haha, if you have any good limits, I would love to see them. So don't hesitate if you find any!
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