## Precalculus (6th Edition) Blitzer

The function f\left( x \right)=\left\{ \begin{align} & x-1\text{ if }x\le 1 \\ & {{x}^{2}}\text{ if }x>1 \end{align} \right. is discontinuous for $x=1$.
Consider the function f\left( x \right)=\left\{ \begin{align} & x-1\text{ if }x\le 1 \\ & {{x}^{2}}\text{ if }x>1 \end{align} \right., Find the value of $f\left( x \right)$ at $x=1$, From the definition of the function, for $x=1$, $f\left( x \right)=x-1$ Then the value of $f\left( x \right)$ at $x=1$ is, \begin{align} & f\left( 1 \right)=1-1 \\ & =0 \end{align} The function is defined at the point $x=1$. Now find the value of $\,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$, First find the left-hand limit of $\,f\left( x \right)$, That is, $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1-1=0$ Now find the right-hand limit of $\,f\left( x \right)$, That is, $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{1}^{2}}=1$ Since, the left-hand limit and right-hand limit are not equal, that is $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$. Thus, $\,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$ does not exist. Thus, the function does not satisfy the second property of being continuous. Hence, the function f\left( x \right)=\left\{ \begin{align} & x-1\text{ if }x\le 1 \\ & {{x}^{2}}\text{ if }x>1 \end{align} \right. is discontinuous at $x=1$.