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Homework Help: Lim of

  1. Sep 5, 2006 #1
    [tex] \lim_{x\rightarrow\infty} tanh x = \lim_{x\rightarrow\infty} \frac{sinh x}{cosh x}[/tex]
    [tex] = \lim_{x\rightarrow\infty} \frac{e^x - e^{-x}}{e^x + e^{-x}}[/tex]
    [tex] = \lim_{x\rightarrow\infty} \frac{e^{2x} -1}{e^{2x} +1}[/tex]

    what now?
     
  2. jcsd
  3. Sep 5, 2006 #2

    AKG

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    Evaluate the limit.
     
  4. Sep 5, 2006 #3

    benorin

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    [tex] =^{H} \lim_{x\rightarrow\infty} \frac{2e^{2x}}{2e^{2x}}=1[/tex]

    where the symbol [tex] =^{H}[/tex] denotes the use of l'Hospital's rule.
     
  5. Sep 5, 2006 #4
    thank you.
     
  6. Sep 6, 2006 #5

    HallsofIvy

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    Swatting a fly with a sledgehammer!
    Multiply both numerator and denominator of
    [tex] = \lim_{x\rightarrow\infty} \frac{e^x - e^{-x}}{e^x + e^{-x}}[/tex]
    by e-x rather than ex and you get
    [tex] = \lim_{x\rightarrow\infty} \frac{1- e^{-2x}}{1+ e^{-2x}}[/tex]
    which is obvious.
     
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