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Math Help - tanhx hyperbolic

  1. #1
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    tanhx hyperbolic

    ex - e-x
    tanh(x) =----------
    ex + e-x



    I know lim tanh (x)=+-1 as x tends to infinity. How do i prove tanh(R) = (-1,1)

    thanks
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  2. #2
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    Quote Originally Posted by charikaar View Post
    ex - e-x
    tanh(x) =----------
    ex + e-x



    I know lim tanh (x)=+-1 as x tends to infinity. How do i prove tanh(R) = (-1,1)

    thanks
    y = 1 and y = -1 are the horizontal asymptotes and the graph lies between these two asymptotes.
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  3. #3
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    Quote Originally Posted by charikaar View Post
    ex - e-x
    tanh(x) =----------
    ex + e-x
    Came out a bit strange didn't it! The internet just doesn't respect initial spaces so it is difficult to "typeset" things like that. LaTex is much better:
    tanh(x)= \frac{e^x- e^{-x}}{e^x+ e^{-x}}
    (click on that to see the code)

    Since exponentials are always positive, we always have the numerator smaller than the denominator, in absolute value. That tells us that the fraction is always between -1 and 1.

    As x goes to infinity, e^{-x} goes to 0 so tanh(x) goes to 1. As x goes to negative infinity, e^x goes to 0 so tanh(x) goes to -1.



    I know lim tanh (x)=+-1 as x tends to infinity. How do i prove tanh(R) = (-1,1)

    thanks
    [/QUOTE]
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