How can we arrive at the approximation $\tanh (x)\approx 1-e^{-2 a}$, for $x^2=a^2+\delta$ with $a>1$ and $\delta\to 0$?

Question

I have seen somewhere that if we have $x^2=a^2+\delta$, where $a>1$ and $\delta\rightarrow0$, then we can use this approximation $$ \tanh (x)\approx 1-e^{-2 a} $$ Can someone explain how we can arrive at this relation?

Answer 1

Note that $\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{1-e^{-2x}}{1+e^{-2x}} \approx 1-e^{-2x}$ for large $x$ and that $\tanh(x)\approx\tanh(a)$ for small $\delta$.

Edit: Taking into account the comment above and below in fact the approximation $1-2e^{-2a}$ seems to be more appropriate. It can be justified by using the first two terms of the Taylor series for $\frac{1-y}{1+y}$ with $y=e^{-2a}$.