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Math Help - Limit Marathon

  1. #16
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by wingless View Post
    Since power series are restricted (by Mathstud) for this one, you can simply multiply \sin x \ln x by \frac{x}{x} like Krizalid did above.
    Yes, but if anyone does any more, you can assert that f(x)\sim{g(x)}\text{ as }x\to{c}

    but only if you show that \lim_{x\to{c}}\frac{f(x)}{g(x)}=1
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  2. #17
    MHF Contributor Mathstud28's Avatar
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    Ok, here is a really really easy one.

    \lim_{x\to\infty}\bigg(a^x+b^x+c^x+d^x+...\bigg)^{  \frac{1}{x}}

    where a>b>c>d>...>0

    EDIT: oops...a typo
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  3. #18
    Super Member wingless's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Ok, here is a really really easy one.

    \lim_{x\to\infty}\bigg(a^x+b^x+c^x+d^x+...\bigg)^{  \frac{1}{x}}

    where a>b>c>d>...>0
    \lim_{x\to\infty}\left (a^x+b^x+c^x+d^x+...\right )^{\frac{1}{x}}

    a^x will grow faster than all other terms. So we can ignore the others and conclude that,

    \lim_{x\to\infty}\left (a^x \right )^{\frac{1}{x}}

    a

    ---

    A more formal approach is here:

    \lim_{x\to\infty}\left (a^x+b^x+c^x+d^x+...\right )^{\frac{1}{x}}

    \lim_{x\to\infty} e^{ \frac{1}{x}\ln(a^x+b^x+c^x+d^x+...)}

    Using L'hopital,

    \lim_{x\to\infty} e^{ \frac{a^x \ln a + b^x \ln b + c^x \ln c + d^x \ln d + ...}{a^x + b^x + c^x + d^x + ...} }

    Now divide all terms by a^x, then all terms except a^x \ln a and a^x will go to zero. So e^{ \frac{a^x \ln a}{a^x} } = e^{\ln a} = a
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  4. #19
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Ok, here is a really really easy one.

    \lim_{x\to\infty}\bigg(a^x+b^x+c^x+d^x+...\bigg)^{  \frac{1}{x}}

    where a>b>c>d>...>0

    EDIT: oops...a typo
    No takers? Oh well,

    There are a couple of ways to do this.

    The smart*** way would be that since

    a>b>c>d>...>0

    we have that a^x+b^x+c^x+d^x+...\sim{a^x}

    This can be shown by the limit

    \lim_{x\to\infty}\frac{a^x+b^x+c^x+d^x+...}{a^x}=\  lim_{x\to\infty}\frac{1+\bigg(\frac{b}{a}\bigg)^x+  \bigg(\frac{c}{a}\bigg)^x+\bigg(\frac{d}{a}\bigg)^  x+...}{1}=1

    this was gotten from the fact that \lim_{x\to\infty}\bigg(\frac{\gamma}{\delta}\bigg)  ^x=0 if \delta>\gamma

    Therefore

    \lim_{x\to\infty}\bigg(a^x+b^x+c^x+d^x+...\bigg)^{  \frac{1}{x}}\sim\lim_{x\to\infty}(a^x)^{\frac{1}{x  }}=a


    The other answer that is better in my opinion would be the use of the ratio/root test trick

    \lim_{x\to\infty}\bigg(a^x+b^x+c^x+d^x\bigg)^{\fra  c{1}{x}}=\lim_{x\to\infty}\frac{a\cdot{a^x}+b\cdot  {b^x}+c\cdot{c^x}+d\cdot{d^x}+...}{a^x+b^x+c^x+d^x  +...}

    Now by termwise division with a^x

    we get

    \lim_{x\to\infty}\frac{a+b\bigg(\frac{b}{a}\bigg)^  x+c\bigg(\frac{c}{a}\bigg)^x+d\bigg(\frac{d}{a}\bi  gg)^x+...}{1+\bigg(\frac{b}{a}\bigg)^x+\bigg(\frac  {c}{a}\bigg)^x+\bigg(\frac{d}{a}\bigg)^x+...}=a
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  5. #20
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by wingless View Post
    \lim_{x\to\infty}\left (a^x+b^x+c^x+d^x+...\right )^{\frac{1}{x}}

    a^x will grow faster than all other terms. So we can ignore the others and conclude that,

    \lim_{x\to\infty}\left (a^x \right )^{\frac{1}{x}}

    a

    ---

    A more formal approach is here:

    \lim_{x\to\infty}\left (a^x+b^x+c^x+d^x+...\right )^{\frac{1}{x}}

    \lim_{x\to\infty} e^{ \frac{1}{x}\ln(a^x+b^x+c^x+d^x+...)}

    Using \color{red}{L'hopital's},

    \lim_{x\to\infty} e^{ \frac{a^x \ln a + b^x \ln b + c^x \ln c + d^x \ln d + ...}{a^x + b^x + c^x + d^x + ...} }

    Now divide all terms by a^x, then all terms except a^x \ln a and a^x will go to zero. So e^{ \frac{a^x \ln a}{a^x} } = e^{\ln a} = a
    Good answer! The only thing is, that you did not show that all the other terms can be ignored via a limit as I did in mine (No big deal, I think we all know it to be true)...and did you just use L'hopital's?


    If anyone sees a problem with any of my solutions let me know
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  6. #21
    Super Member wingless's Avatar
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    Well, my first answer is the same as yours, I just didn't show the details. And if you don't want to evaluate the limit using \sim, use the L'hopital solution.. =)
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  7. #22
    MHF Contributor Mathstud28's Avatar
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    EDITsorry, I am making these up as we go) Tis is another easy one, if someone would like to post another one I would be more than happy to give up the posting position

    Find the smallest integer \alpha such that

    \lim_{n\to\infty}\frac{(n!)^{\alpha}}{(3n)!}>1
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  8. #23
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    EDITsorry, I am making these up as we go) Tis is another easy one, if someone would like to post another one I would be more than happy to give up the posting position

    Find the smallest integer \alpha such that

    \lim_{n\to\infty}\frac{(n!)^{\alpha}}{(3n)!}>1
    Here is a hint

    So expanding we get

    \lim_{n\to\infty}\frac{(2\pi{n})^{\frac{\alpha}{2}  }(n^n)^{\alpha}(e^{-n})^{\alpha}}{\sqrt{6\pi{n}}27^n(n^n)^3(e^{-n})^3}
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  9. #24
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Here is a hint

    So expanding we get

    \lim_{n\to\infty}\frac{(2\pi{n})^{\frac{\alpha}{2}  }(n^n)^{\alpha}(e^{-n})^{\alpha}}{\sqrt{6\pi{n}}27^n(n^n)^3(e^{-n})^3}
    Thank the Heaven's for Stirling's approximation. Some days it seems to me we wouldn't have Physics without it.

    -Dan
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  10. #25
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Here is a hint

    So expanding we get

    \lim_{n\to\infty}\frac{(2\pi{n})^{\frac{\alpha}{2}  }(n^n)^{\alpha}(e^{-n})^{\alpha}}{\sqrt{6\pi{n}}27^n(n^n)^3(e^{-n})^3}
    Ok, since no one is answering.

    It cannot be one or two, since that would leave a factor of n^n and \alpha=3 would still leave that pesky 27^n in the denominator, but \alpha=4 does the trick

    Does no one have any hard limits?
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  11. #26
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Ok, since no one is answering.

    It cannot be one or two, since that would leave a factor of n^n and \alpha=3 would still leave that pesky 27^n in the denominator, but \alpha=4 does the trick
    Not that you are incorrect in any way about this, but you are aware that this makes the limit go from 0 ( \alpha = 3) to \infty ( \alpha = 4)?

    I wonder if there is a way to find an \alpha such that the limit is equal to 1?

    -Dan
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  12. #27
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by topsquark View Post
    Not that you are incorrect in any way about this, but you are aware that this makes the limit go from 0 ( \alpha = 3) to \infty ( \alpha = 4)?

    I wonder if there is a way to find an \alpha such that the limit is equal to 1?

    -Dan
    I do not think so, for we would have to find a value such that a function of the form a^n and a function n^{b} are equal as n goes to infinity, which would be impossible since n^b\prec{a^n}
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  13. #28
    Eater of Worlds
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    Try these:

    \lim_{y\to\rightarrow{0}}\frac{1}{y}\int_{0}^{\pi}  tan(ysin(x))dx

    \lim_{n\to\rightarrow{\infty}}\frac{1}{n}\sum_{j=1  }^{n}\sum_{k=1}^{n}\frac{j}{j^{2}+k^{2}}
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  14. #29
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Try these:

    \lim_{y\to\rightarrow{0}}\frac{1}{y}\int_{0}^{\pi}  tan(ysin(x))dx

    \lim_{n\to\rightarrow{\infty}}\frac{1}{n}\sum_{j=1  }^{n}\sum_{k=1}^{n}\frac{j}{j^{2}+k^{2}}
    Is the first one two?

    \lim_{y\to{0}}\frac{\int_0^{\pi}\tan(y\sin(x))dx}{  y}=\int_0^{\pi}\sin(x)dx=2?
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  15. #30
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Is the first one two?

    \lim_{y\to{0}}\frac{\int_0^{\pi}\tan(y\sin(x))dx}{  y}=\int_0^{\pi}\sin(x)dx=2?
    It would appear that Mathstud28 is doing another series expansion, but didn't tell us!

    -Dan
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