## Precalculus (6th Edition) Blitzer

The function f\left( x \right)=\left\{ \begin{align} & x-5\text{ if }x\le 0 \\ & {{x}^{2}}+x-5\text{ if }x>0 \end{align} \right. is continuous at $0$.
Consider the function f\left( x \right)=\left\{ \begin{align} & x-5\text{ if }x\le 0 \\ & {{x}^{2}}+x-5\text{ if }x>0 \end{align} \right., Find the value of $f\left( x \right)$ at $a=0$, \begin{align} & f\left( 0 \right)=0-5 \\ & =-5 \end{align} The function is defined at the point $a=0$. Now find the value of $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$. First, find the left hand limit of $\,f\left( x \right)$. The function $f\left( x \right)=x-5$ when $x<0$. $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0-5=-5$ Now find the right hand limit of $\,f\left( x \right)$. The function $f\left( x \right)={{x}^{2}}+x-5$ when $x>0$. $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{0}^{2}}+0-5=-5$ Since the left hand limit and right hand limit are equal, that is $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=-5=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ Thus, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=-5$ From the above steps, $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=-5=f\left( 0 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function f\left( x \right)=\left\{ \begin{align} & x-5\text{ if }x\le 0 \\ & {{x}^{2}}+x-5\text{ if }x>0 \end{align} \right. is continuous at $0$.