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Approximation of a hyperbolic function

  1. May 5, 2017 #1
    1. The problem statement, all variables and given/known data
    Hy guys I am having an issue with approximating this first question, which I have shown below.


    Now my problem is not so much solving it but I have been thinking that if given the same question without knowing that it approximates to so for example the question I am thinking of is: approximate the function: ##\frac{cosh x}{sinh x}-1##.

    2. Relevant equations

    3. The attempt at a solution

    So alls I have at the moment is ## \lim_{x\to\infty}\frac{2e^{-2x}}{1-2e^{-2x}}##
    and this is where I get stuck how would I use limits to show the it approximates to what they have given. I was thinking of just taking the limit of the bottom line, and this would give the ans but I am not 100% sure that you can actually do that, another way I was thinking is showing what the end of the function dose, but then I would have to dived through and I am not sure that the function can be divided. Could someone please give me some advice much appreciated.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 5, 2017 #2
    I think you probably use @Taylor_1989 series and inequalities.
  4. May 5, 2017 #3


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    There's an error in the denominator of

  5. May 5, 2017 #4


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    So we have 2*cosh(x) = [e^x + e^-x] and 2*sinh(x) = [e^x - e^-x]. You would agree that cosh(x) / sinh(x) is equal to (2*cosh) / (2*sinh). Now replace the 1 with (2*sinh) / (2*sinh), so you now have : [e^x + e^-x - (e^x - e^-x)] / [e^x - e^-x]. With a little rearranging, you may be able to simplify it.
  6. May 5, 2017 #5


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    Or actually with what you have, the denominator of 1 - 2*e^(-2x) approaches 1 as x gets large. So you can say it is equal to 2*e^(-2x) in the numerator, and approximately 1 in the denominator, so it approaches 2*e^(-2x)
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