Why is $\lvert \tanh x\rvert < 1$? In the solutions to my exercises, it says that it is easy to see that $\lvert \tanh x\rvert < 1$. Sadly, this doesn't really help me. I know that $$\lvert \tanh x\rvert = \left\lvert \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \right\rvert = \left\lvert \frac{1-e^{-2x}}{1 + e^{-2x}} \right\rvert$$
However, I don't see how this would imply or even help me show $\lvert \tanh x\rvert < 1$.
 A: $$1-\tanh^2x=1-\left(\dfrac{e^x-e^{-x}}{e^x+e^{-x}}\right)^2=\dfrac4{(e^x+e^{-x})^2}>0$$
A: $$
\lvert \tanh x\rvert =   \frac{\lvert 1-e^{-2x} \rvert}{|1 + e^{-2x}|} 
$$
The exponential term $ e^{-2x}$ is always positive, so the numerator is always less than the denominator, namely $\lvert 1-e^{-2x} \rvert < |1 + e^{-2x}|$. This implies $\lvert \tanh x\rvert <1$ .
A: Put $t = e^x + e^{-x}$. Then$$ (\tanh x)^2 = \left( \frac{e^x - e^{-x}}{e^x + e^{-x}}\right)^2 = \frac{ t^2 - 4} {t^2} = 1  - \frac{4}{t^2} $$
with $t>0$.
A: Note that $e^x - e^{-x} < e^x$ since $e^{-x} > 0$. But then
$$ \frac{e^x - e^{-x}}{e^x + e^{-x}} < \frac{e^x}{e^x + e^{-x}}.$$
However $e^x + e^{-x} > e^x$, so $\dfrac{1}{e^x + e^{-x}} < \dfrac{1}{e^x}$ and
$$ \frac{e^x}{e^x + e^{-x}} < \frac{e^x}{e^x} = 1.$$
Thus
$$ \frac{e^x - e^{-x}}{e^x + e^{-x}} < 1.$$
The lower bound follows very similar arguments and I'll leave that to you to piece together.
A: $$\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}.$$

*

*As $x\rightarrow-\infty,\; e^{2x}\rightarrow0$ so
$\displaystyle\tanh(x)=\frac{e^{2x}-1}{e^{2x}+1}\rightarrow-1.$

*As $x$ increases, $1-e^{-2x}$ increases while $1+e^{-2x}$ decreases;
so, $\tanh(x)=\displaystyle\frac{1-e^{-2x}}{1+e^{-2x}}$ is an
increasing function.

*As $x\rightarrow\infty,\; e^{-2x}\rightarrow0$ so
$\displaystyle\tanh(x)=\frac{1-e^{-2x}}{1+e^{-2x}}\rightarrow1.$
Therefore $\displaystyle-1<\tanh(x)<1.$
