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Homework Help: Finding hypberbolic/exponential limit

  1. Mar 9, 2008 #1
    [SOLVED] finding hypberbolic/exponential limit

    1. The problem statement, all variables and given/known data
    [tex]\mathop{\lim} \limits_{x \to \infty} \frac{cosh(2x)}{e^2^x}[/tex]

    2. Relevant equations
    [tex]cosh = \frac{e^x+e^-^x}{2}[/tex]

    3. The attempt at a solution
    [tex]\mathop{\lim} \limits_{x \to \infty} \frac{cosh(2x)}{e^2^x} \rightarrow \frac{\frac{e^2^x+e^-^2^x}{2}}{e^2^x} \rightarrow \frac{e^2^x+e^-^2^x}{2}*\frac{1}{e^2^x} \rightarrow \frac{\ln}{\ln} (\frac{e^2^x+e^-^2^x}{2e^2^x}) \rightarrow \frac{2x}{2x\ln2} - \frac{2x}{2x\ln2} = 0[/tex]

    Hey, it's me again. This latex script is groovy. Am I on the right track here? Cheers.
  2. jcsd
  3. Mar 9, 2008 #2


    [tex]\frac 1 2\lim_{x\rightarrow\infty}\left(\frac{e^{2x}}{e^{2x}}+\frac{e^{-2x}}{e^{2x}}\right)[/tex]

    Take it from here ...
    Last edited: Mar 9, 2008
  4. Mar 9, 2008 #3


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    Science Advisor

    I have no idea what
    [tex]\frac{\ln}{\ln} (\frac{e^2^x+e^-^2^x}{2e^2^x}) \[/tex]
  5. Mar 9, 2008 #4
    Thanks for the replies.

    I thought I should take the ln of the fraction to eliminate the exponents. However, rocophysics' reply indicates this strategy was incorrect. Unfortunately, I am still confused.

    On account of the cosh(2x) in the original problem, I believe steps 2 and 3 quoted above should read as follows:


    [tex]\frac 1 2\lim_{x\rightarrow\infty}\left(\frac{e^2^x}{e^{2x}}+\frac{e^{-2x}}{e^{2x}}\right)[/tex]

    and then maybe...

    [tex]\frac 1 2\lim_{x\rightarrow\infty}\left(2\right) = 1[/tex]

  6. Mar 9, 2008 #5
    SHOOT!!! So sorry :) Thank you for fixing it ... I was so caught up on the LaTeX, lol.

    So your answer is 1/2
  7. Mar 9, 2008 #6
    Hmmm. If [tex]\left(\frac{e^2^x}{e^{2x}}+\frac{e^{-2x}}{e^{2x}}\right)=2[/tex] shouldn't the answer be 1?

  8. Mar 9, 2008 #7
    I pulled out the 1/2 infront of the limit.

    [tex]\frac 1 2\lim_{x\rightarrow\infty}\left(\frac{e^{2x}}{e^{2 x}}+\frac{e^{-2x}}{e^{2x}}\right)[/tex]

    [tex]\frac 1 2\lim_{x\rightarrow\infty}\left(1+\frac{1}{e^{4x}}\right)[/tex]

    [tex]\frac 1 2\lim_{x\rightarrow\infty}\left(1+0\right)[/tex]
  9. Mar 9, 2008 #8
    Aha! I've got it now. Thanks.
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