585. Find the one-sided limits of the function

$$
f(x)=\operatorname{arctg} \frac{1}{x-1}(x \neq 1) \text { as } x \rightarrow 1
$$
Solution. $f(1+0)=\lim _{\substack{x \rightarrow 1+0 \\(x>1)}} \operatorname{arctg} \frac{1}{x-1}=\lim _{z \rightarrow+\infty} \operatorname{arctg} z=\frac{\pi}{2}$, where $z=\frac{1}{x-1}$; from this it is clear that if $x \rightarrow 1+0$, then $z \rightarrow+\infty$.

$$
f(1-0)=\lim _{\substack{x \rightarrow 1-0 \\(x<1)}} \operatorname{arctg} \frac{1}{x-1}=\lim _{z \rightarrow-\infty} \operatorname{arctg} z=-\frac{\pi}{2}
$$

The limits turned out to be unequal, therefore, the limit of the function at the point $a=1$ does not exist.