Answer
$\displaystyle \lim_{x\rightarrow 0}\coth x$ does not exist.
Work Step by Step
$\displaystyle \coth x=\frac{1}{\tanh x}=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}},\quad x\neq 0$
As x aproaches zero from the left, $x\rightarrow 0^{-}$,
In the numerator, $e^{x}+e^{-x}$ there is a positive number, approaching 2
and in the denominator, $e^{x}$ is slightly less than 1, and $e^{-x}$ is slighty more than 1, so
the denominator is negative as it approaches zero.
$\displaystyle \lim_{x\rightarrow 0^{-}}\coth x =-\infty$
As x aproaches zero from the right, $x\rightarrow 0^{+}$,
$e^{x}$ is slightly more than 1, and $e^{-x}$ is slighty less than 1, so
the denominator is positive as it approaches zero. The numerator approaches $+2.$
$\displaystyle \lim_{x\rightarrow 0^{+}}\coth x =+\infty$
$\displaystyle \lim_{x\rightarrow 0}\coth x$ does not exist.
