Answer
$0$
Work Step by Step
$y=\csc^{-1}x$ is the number in $[-\pi/2,0) \cup(0, \pi/2]$ for which $\csc y=x$
$($In terms of sine, $\displaystyle \sin y=\frac{1}{x}$)
We want an angle for which sine approaches 0 from the left, so the cosecant $\rightarrow\infty$. This angle (in radians) is $0$, (we observe $[-\pi/2,0) \cup(0, \pi/2]$).
Alternatively (if in doubt), we can reach the same conclusion by observing the graph of $y=\csc x$, (also written as $\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{s}\mathrm{c} x$) when $ x\rightarrow-\infty$. See below.
