## Precalculus (6th Edition) Blitzer

The number for which the function f\left( x \right)=\left\{ \begin{align} & x+6\text{ if }x\le 0 \\ & \text{6 if }02 \end{align} \right. is discontinuous at $x=2$.
Consider the function f\left( x \right)=\left\{ \begin{align} & x+6\text{ if }x\le 0 \\ & \text{6 if }02 \end{align} \right., First check the discontinuity of the function at $x=0$ Find the value of $f\left( x \right)$ at $x=0$, From the definition of the function, for $x=0$, $f\left( x \right)=x+6$ Then the value of $f\left( x \right)$ at $x=0$ is, \begin{align} & f\left( 0 \right)=0+6 \\ & =6 \end{align} The function is defined at the point $x=0$. Now find the value of $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$, First find the left-hand limit of $\,f\left( x \right)$, That is, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0+6=6$ Now find the right-hand limit of $\,f\left( x \right)$, That is, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=6$ Since the left-hand limit and right-hand limit are equal, that is $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=6=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ Thus $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=6$. From the above steps, $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=6=f\left( 0 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function f\left( x \right)=\left\{ \begin{align} & x+6\text{ if }x\le 0 \\ & \text{6 if }02 \end{align} \right. is continuous at $x=0$. Now check the discontinuity of the function at $x=2$ Find the value of $f\left( x \right)$ at $x=2$, From the definition of the function, $f\left( 2 \right)=6$ The function is defined at the point $x=2$. Now find the value of $\,\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$, First find the left-hand limit of $\,f\left( x \right)$, That is, $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=6$ Now find the right-hand limit of $\,f\left( x \right)$, That is, $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{2}^{2}}+1=5$ Since the left-hand limit and right-hand limit are not equal, that is $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ Thus, $\,\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$ do 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+6\text{ if }x\le 0 \\ & \text{6 if }02 \end{align} \right. is discontinuous at $x=2$. Thus, the number for which the function f\left( x \right)=\left\{ \begin{align} & x+6\text{ if }x\le 0 \\ & \text{6 if }02 \end{align} \right. is discontinuous at $x=2$.