15 December 2021

Statement

Is the following function continuous at \(x = 1\) ?

\(f(x) = \begin{cases} \frac{1}{x-1} & x \neq 1 \\ 2 & x=1 \end{cases}\)

Solution

\(f(x) = \begin{cases} \frac{1}{x-1} & x \neq 1 \\ 2 & x=1 \end{cases}\)

Let's start by finding the limit of \(f(x)\) when \(x \to 1\) from both sides.


From the right side

\(\lim\limits_{x \to 1^{+}} f(x)\)

\(= \lim\limits_{x \to 1^{+}} \frac{1}{x-1}\)

\(= \lim\limits_{x \to 1^{+}} \frac{1}{1^{+}-1}\)

\(= \lim\limits_{x \to 1^{+}} \frac{1}{0^{+}}\)

\(= +\infty\)


From the left side

\(\lim\limits_{x \to 1^{-}} f(x)\)

\(= \lim\limits_{x \to 1^{-}} \frac{1}{x-1}\)

\(= \lim\limits_{x \to 1^{-}} \frac{1}{1^{-}-1}\)

\(= \lim\limits_{x \to 1^{-}} \frac{1}{0^{-}}\)

\(= -\infty\)

This already means there is a discontinuity to infinity at \(x=1\). And since \(f(1) = 2\) there are double the reasons to believe the discontinuity exists.