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
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:
$\displaystyle tanh(x)= \frac{e^x- e^{-x}}{e^x+ e^{-x}}$
(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, $\displaystyle e^{-x}$ goes to 0 so tanh(x) goes to 1. As x goes to negative infinity, $\displaystyle e^x$ 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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