Precise value of $\coth(x)$ for large values of $x$? We know that for large values of $x$, $\cosh (x)\approx \sinh (x)$, and then $\lim_{x\to \infty } \, \coth (x)=1$. My question is that for example is the result of $\coth (100)$ exactly one? Or it is an estimated value?
${\cosh(100)}$ will actually be a very large number. Did you mean ${\coth(100)}$? If so, no, it will not be exactly $1$, it's estimated. But it will be very, very close to it. The fact the limit is $1$ as ${x\rightarrow\infty}$ just means that they behave &quot;asymptotically&quot; the same as ${x}$ get's larger.
Large $x$, or $x\gg1$, means $e^{-x}\ll1$. Then, $\coth x $ can be expressed as
$$\coth x =\frac{e^x+e^{-x}}{e^x-e^{-x}}
= \frac{1+e^{-2x}}{1-e^{-2x}}\approx(1+e^{-2x})^2\approx1+2e^{-2x}
$$
For example, $\coth 100 \approx 1+2e^{-200}$.