$\begin{aligned} & \lim _{x \rightarrow 1-} \frac{1}{|1-x|}=\lim _{h \rightarrow 0} \frac{1}{1-(1-h)}=\infty \\ & \text { and } \lim _{x \rightarrow 1+} \frac{1}{|1-x|}=\lim _{h \rightarrow 0} \frac{1}{1+h-1}=\infty \\ & \text { Hence } \lim _{x \rightarrow 1 \mid} \frac{1}{1-x \mid}=\infty .\end{aligned}$