Thread: 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

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.

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:

(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, goes to 0 so tanh(x) goes to 1. As x goes to negative infinity, 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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