Answer
$\displaystyle \lim_{x\rightarrow-\infty} \mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h} x=0$
Work Step by Step
$\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}\ x=\displaystyle \frac{1}{\sinh x}=\frac{2}{e^{x}-e^{-x}},\quad x\neq 0$
When $ x\rightarrow-\infty,\qquad e^{x}\rightarrow 0\quad e^{-x}\rightarrow\infty$
the numerator is constant, the denominator $\rightarrow\infty$
$\displaystyle \lim_{x\rightarrow-\infty} \mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h} x=0$
