# tanhx hyperbolic

• Feb 6th 2010, 10:48 AM
charikaar
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
• Feb 6th 2010, 01:58 PM
mr fantastic
Quote:

Originally Posted by charikaar
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.
• Feb 7th 2010, 01:27 AM
HallsofIvy
Quote:

Originally Posted by charikaar
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
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