Why is $\lvert \tanh x\rvert < 1$?

Question

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 $\lvert \tanh x\rvert < 1$.

Answer 1

$$1-\tanh^2x=1-\left(\dfrac{e^x-e^{-x}}{e^x+e^{-x}}\right)^2=\dfrac4{(e^x+e^{-x})^2}>0$$

Answer 2

$$ \lvert \tanh x\rvert = \frac{\lvert 1-e^{-2x} \rvert}{|1 + e^{-2x}|} $$ 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 $\lvert \tanh x\rvert <1$ .

Answer 3

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 $t>0$.

Answer 4

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.

Answer 5

$$\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}.$$

• 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 $\displaystyle-1<\tanh(x)<1.$