## Calculus, 10th Edition (Anton)

$$f(x)=\left\{\begin{array}{ll}{\frac{3}{x-1},} & {x \neq 1} \\ {3,} & {x=1}\end{array}\right.$$ value $x=1$ , at which $f$ is not continuous
$$f(x)=\left\{\begin{array}{ll}{\frac{3}{x-1},} & {x \neq 1} \\ {3,} & {x=1}\end{array}\right.$$ We obtain that $$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} \frac{3}{x-1}= \infty$$ $$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow-1^{-}} \frac{3}{x-1}=-\infty$$ $$\lim _{x \rightarrow 1^{-}} f(x) \ne \lim _{x \rightarrow 1^{-}} f(x)$$ therefore $$\lim _{x \rightarrow 1} f(x)$$ does not exist therefore at $x=1$ , $f$ is not continuous