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Limit of function

  1. Dec 20, 2014 #1
    • Member warned about not using the homework template

    I'm trying to find the limit as x tends to zero of the function (tanhx-x)/x
    this is what i have but i have no idea if i am on the right lines?
    lim x-> 0 can be split up in to two problems:
    limx->0 (tanhx)/x - limx->0 x/x
    limx->0 x/x = 1
    limx->0 (tanhx)/x can be expressed as
    limx->0 ((e^x-e^-x)/e^x+e^-x))/x
    limx->0 (x(e^x-e^-x)/e^x+e^-x)) =0
    which leave the limit as -1 but wolfram gives a limit of 0

    how should i be approaching this problem?
    many thanks
  2. jcsd
  3. Dec 20, 2014 #2


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    in your final expression, I'm not sure how you're getting -1
    [e^0 - e^(-0)] =? I think you plugged in your limit wrongly.

    Edit: Oh wait, I thought you were saying that the limit of tanh was -1.
    Your problem lies in the fact that ##\frac{tanh(x)}{x} \neq \frac{x(e^x-e^{-x}}{e^x +e^{-x})}##

    ##\frac{tanh(x)}{x} = \frac{e^x-e^{-x}}{x(e^x +e^{-x})}##
    Last edited: Dec 20, 2014
  4. Dec 20, 2014 #3
    With the x/x you can use l'hopitals rule to get round the fact that you have somepthing that is tending to 0/0 but for the tanh(x)/x part you cant? so how do you evaluate it? and im a bit confused as to how
    Because i thought
    ((e^x-e^-x)/e^x+e^-x))/x was the same as:
    ((e^x-e^-x)/e^x+e^-x)) times x/1 ?

    many thanks for your reply :)
  5. Dec 20, 2014 #4


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    You can use the hospital rule. But you didn't express tanh x/x correctly.
    ##\frac{tanh x}{x} = \frac{1}{x}tanh x = \frac{1}{x}\frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{e^x-e^{-x}}{x(e^x + e^{-x})}##
  6. Dec 20, 2014 #5


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    In your case, you end up with a 0/1, but in this case you don't.
  7. Dec 20, 2014 #6
    ah brilliant much clearer thanks! :)
  8. Dec 20, 2014 #7


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    Not a problem.
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