15 December 2021

Statement

Is the following function continuous at \(x = 2\) ?

\(f(x) = \begin{cases} 1 - x & x \leq 2 \\ x^2 - 2x & x > 2 \end{cases}\)

Solution

\(f(x) = \begin{cases} 1 - x & x \leq 2 \\ x^2 - 2x & x > 2 \end{cases}\)

Let's start by finding the limit of \(f(x)\) when \(x \to 2\) from both sides.


From the right side

\(\lim\limits_{x \to 2^{+}} f(x)\)

\(= \lim\limits_{x \to 2^{+}} x^2 - 2x\)

\(= \lim\limits_{x \to 2^{+}} 2^2 - 4\)

\(= 0\)


From the left side

\(\lim\limits_{x \to 2^{-}} f(x)\)

\(= \lim\limits_{x \to 2^{-}} 1 - x\)

\(= \lim\limits_{x \to 2^{-}} 1 - 2\)

\(= -1\)

This means that \(f(x)\) has a jump at \(x = 2\) since \(\lim\limits_{x \to 2^{-}} f(x) \neq \lim\limits_{x \to 2^{+}} f(x)\) therefore the function is not continuous at \(x = 2\).