Find $\lim\limits_{x\rightarrow 2} f(x)$, if it exists, if $$f( x) =\left\{ \begin{array}{l@{\qquad}ll} 3x+1 & \hbox{if }\quad x\lt 2 \\[3pt] 2x( x-1) & \hbox{if }\quad x\geq 2 \end{array} \right.$$

Figure 19 $$f( x) =\left\{ \begin{array}{l@{\qquad}ll} 3x+1 & \hbox{if } x\lt 2 \\[3pt] 2x( x-1) & \hbox{if } x\geq 2 \end{array} \right.$$

Solution Since the rule for $f$ changes at $2,$ we need to find the one-sided limits of $f$ as $x$ approaches $2$.

For $x\lt 2,$ we use the left-hand limit. $$\begin{equation*} \lim\limits_{x\rightarrow 2^{-}}f(x) =\lim\limits_{x\rightarrow 2^{-}}(3x+1) =\lim\limits_{x\rightarrow 2^{-}}(3x) +\lim\limits_{x\rightarrow 2^{-}}1=3\!\lim\limits_{x\rightarrow 2^{-}}x+1=3(2) +1=7 \end{equation*}$$

For $x>2,$ we use the right-hand limit. $$\begin{eqnarray*} \lim\limits_{x\rightarrow 2^{+}}f(x) &=&\lim\limits_{x\rightarrow 2^{+}}[ 2x(x-1)] =\lim\limits_{x\rightarrow 2^{+}}(2x) \cdot \lim\limits_{x\rightarrow 2^{+}}(x-1)\\[4pt] &=&2\lim\limits_{x\rightarrow 2^{+}}x\cdot \Big[ \lim\limits_{x\rightarrow 2^{+}}x-\lim\limits_{x\rightarrow 2^{+}}1\Big] =2\cdot 2\,[2-1] =4 \end{eqnarray*}$$

Since $\lim\limits_{x\rightarrow 2^{-}}f(x) =7\neq \lim\limits_{x\rightarrow 2^{+}}f(x) =4,$ $\lim\limits_{x \rightarrow 2}f(x)$ does not exist.