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Homework Help: Limit problem

  1. Nov 27, 2012 #1
    Here it is:
    \lim_{x\to\infty} tanh(x)=\lim_{x\to\infty} sinh(x)/cosh(x)=\lim_{x\to\infty} \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}

    Can you just assume that the [tex]e^{-x}[/tex] will approach 0 and the [tex]e^{x}[/tex] will cancel and the limit will reduce to 1?
  2. jcsd
  3. Nov 27, 2012 #2
    Re: Limit

    Well, it's correct. But perhaps your teacher might want a bit more formal and rigorous solution? Depends on your teacher of course.
  4. Nov 27, 2012 #3
    Re: Limit

    That's what I was wondering, what might be a more rigorous approach to this?
  5. Nov 27, 2012 #4
    Re: Limit

    By the way, I'm not sure why this was moved because it's not homework

    Nonetheless, I found a way to do this:

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

    The second limit is simplified enough that we know that it is equivalent to 0.

    [tex]lim_{x\to\infty} \frac{e^{x}}{e^{x}+e^{-x}}[/tex]
    [tex]lim_{x\to\infty} \frac{1}{\frac{e^{x}+e^{-x}}{e^{x}}}[/tex]
    [tex]lim_{x\to\infty} \frac{1}{1+e^{-2x}}[/tex]

    This is enough to show that the first limit is equivalent to 1.
    Last edited: Nov 27, 2012
  6. Nov 27, 2012 #5


    Staff: Mentor

    Re: Limit

    It's much simpler just to factor ex out of the expressions in the numerator and denominator.

    $$ \lim_{x\to\infty} \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$$
    $$ = \lim_{x\to\infty} \frac{e^x(1 - e^{-2x})}{e^x(1 + e^{-2x})} $$

    Also, you should connect your limit expressions with =.
    Fixed the LaTeX in the one below.
  7. Nov 27, 2012 #6
    Re: Limit

    How about trying de L'Hopital's Rule, after you've multiplied the numerator and denominator by ex?
    \lim\frac{ e^x-e^{-x} }{ e^x+e^{-x} } =
    \lim\frac{ e^{2x}-1 }{ e^{2x}+1 } =
    \lim\frac{ 2e^{2x} }{ 2e^{2x} } =
    \lim1 = 1.
  8. Nov 27, 2012 #7
    Re: Limit

    That doesn't matter. It's a homework-style question from a textbook, so it belongs in the homework forums. This is regardless of that it actually is homework or is self-study.
  9. Nov 27, 2012 #8
    Re: Limit

    [tex]That\space would\space have\space saved\space me\space a\space lot\space of\space time...[/tex]
  10. Nov 27, 2012 #9


    Staff: Mentor

    Re: Limit

    As would have factoring ex out of the top and bottom.

    Also, why did you write the text above as LaTeX? LaTeX is very useful for depicting fractions and integrals and such, but some posts with large amounts of LaTeX take a long time to render in some browsers.
  11. Nov 27, 2012 #10
    Re: Limit

    Because Michael was helping me with latex before
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