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?
 A: ${\cosh(100)}$ 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 "asymptotically" the same as ${x}$ get's larger.
A: $\coth(100)\approx{}{}{}{}{}{}{}{}{}{}{}$ 
A: The precise value of $\coth(x)$ for all $x$ is, by definition,
$$\coth(x)\equiv\frac{e^x+e^{-x}}{e^x-e^{-x}}$$
So
$$\coth(100)=\frac{e^{100}+e^{-100}}{e^{100}-e^{-100}}$$
and according to Wolfram Alpha,
$$\coth(100)\approx1.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000276779305347347506129736291$$
So, is not exactly $1$, but it is very, very close to $1$. In general, for "large" $x$, $\coth(x)\approx1$.
A: \begin{align}
& \frac{e^x + e^{-x}}{e^x-e^{-x}} = 1 \\[10pt]
& e^x + e^{-x} = e^x - e^{-x} \\[10pt]
& e^{-x} = -e^{-x} \\[10pt]
& e^{-x} = 0. \qquad \longleftarrow \text{This never happens.} \\[10pt]
& e^{-x} \approx0. \qquad \longleftarrow \text{This happens if $x$ is big.}
\end{align}
A: Let $x=\log(y)$ and consider
$$\coth(x)=\frac{y^2+1}{y^2-1}=1+2\sum_{n=1}^\infty \frac 1 {y^{2n}}=1+2 \sum_{n=1}^\infty e^{-2nx}$$ For $x=100$, the first term is
$$1+2e^{-200}=1+2.76779 \times 10^{-87}$$ which makes a bunch of zeros after the decimal point.
A: 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, $\coth 100 \approx 1+2e^{-200}$.
