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Math Help - Hyperbolic Limit Question

  1. #1
    Kep
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    Hyperbolic Limit Question

    I need to show that \frac {\cosh(w)} {w^3} goes to zero as w goes to infinity. I don't need a rigorous epsilon delta proof. Any help would be appreciated!
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    Quote Originally Posted by Kep View Post
    I need to show that \frac {\cosh(w)} {w^3} goes to zero as w goes to infinity. I don't need a rigorous epsilon delta proof. Any help would be appreciated!
    The limit is not zero. The limit is actually +oo.

    One proof of this limit is to divide the Maclaurin series for cosh(w) by w^3 and then consider the limit. Another proof is to use l'Hopital's Rule since it's an indeterminant form oo/oo. A simple minded approach is to substitute larger and larger values of w into the expression and observe the pattern.
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    Kep
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    That's what I had thought... but Wolfram Alpha seems to think that it is zero...

    Limit of cosh(w)/w^3 as w goes to infinity - Wolfram|Alpha
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    Using L'Hospital's Rule:

    \lim_{w \to \infty}\frac{\cosh{w}}{w^3} = \lim_{w \to \infty}\frac{\sinh{w}}{3w^2}

    Since this is still of the indeterminate form \frac{\infty}{\infty}, use L'Hospital's rule again...

    \lim_{w \to \infty}\frac{\sinh{w}}{3w^2} = \lim_{w \to \infty}\frac{\cosh{w}}{6w}.

    This is still \frac{\infty}{\infty}, so use L'Hospital again.

    \lim_{w \to \infty}\frac{\cosh{w}}{6w} = \lim_{w \to \infty}\frac{\sinh{w}}{6}.


    This clearly tends to \infty as w \to \infty.
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    Quote Originally Posted by Kep View Post
    That's what I had thought... but Wolfram Alpha seems to think that it is zero...

    Limit of cosh(w)/w^3 as w goes to infinity - Wolfram|Alpha
    Yes, I noticed the same thing. There is a moral to the story ....
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    It is interesting that if you click the "show steps" button on the wolfram site, there are no steps, just a repeat of the claim that the limit is 0!
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    Kep
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    Wolfram Alpha: The limit is zero

    Me: Really? Why? *Clicks show steps*

    Wolfram Alpha: The limit IS zero dammit, stop being so curious...



    This is bad news for me though

    I'm supposed to find an upper bound for \mod( \frac {\cot(z)} {z^4}) on the square with vertices ((N+\frac {1} {2})\pi)(\pm 1 \pm i)

    This is then supposed to vanish when multiplied by the perimeter of the square as N goes to infinity.

    My upper bound times perimeter was of that form and since Wolfram Alpha was encouraging I assumed I had got it right...

    Serves me right for not posting the whole question...

    Any ideas? Sorry this might be a bit tough...

    My first attempt was to write cot in terms of exp and use the triangle inequality...
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Kep View Post
    Wolfram Alpha: The limit is zero

    Me: Really? Why? *Clicks show steps*

    Wolfram Alpha: The limit IS zero dammit, stop being so curious...



    This is bad news for me though

    I'm supposed to find an upper bound for \mod( \frac {\cot(z)} {z^4}) on the square with vertices ((N+\frac {1} {2})\pi)(\pm 1 \pm i)

    This is then supposed to vanish when multiplied by the perimeter of the square as N goes to infinity.

    My upper bound times perimeter was of that form and since Wolfram Alpha was encouraging I assumed I had got it right...

    Serves me right for not posting the whole question...

    Any ideas? Sorry this might be a bit tough...

    My first attempt was to write cot in terms of exp and use the triangle inequality...
    What is mod? Modulus?
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  9. #9
    Kep
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    Yeah, mod is modulus. As in \sqrt(x^2+y^2) where x=\Re(\frac {\cot(z)} {z^4}) and y=\Im(\frac {\cot(z)} {z^4})
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