## Precalculus (6th Edition) Blitzer

Consider the piecewise function defined by f\left( x \right)=\left\{ \begin{align} & 1-x\text{ if }x<1 \\ & \text{0 if }x=1\text{ } \\ & {{x}^{2}}-1\text{ if }x>1 \end{align} \right.. We find $\,\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ using $f\left( x \right)=$ $1-x$. We find $\,\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ using $f\left( x \right)=$ ${{x}^{2}}-1$. The function’s definition indicates that $f\left( 1 \right)=$ $0$. If we determine that $\,\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 1 \right)$, we can conclude that the function is continuous at $1$.
Consider the function, f\left( x \right)=\left\{ \begin{align} & 1-x\text{ if }x<1 \\ & \text{0 if }x=1\text{ } \\ & {{x}^{2}}-1\text{ if }x>1 \end{align} \right. The value $\,\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is the value of the function as x nears $1$ from the left. As x nears $1$ from the left, the value of the function is $f\left( x \right)=1-x$. The value $\,\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is the value of the function as x nears $1$ from the right. As x nears $1$ from the right, the value of the function is $f\left( x \right)={{x}^{2}}-1$ From the definition of the function, the value of $f\left( 1 \right)=0$ If $\,\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 1 \right)$, then the function satisfies all the conditions of being continuous, so the function is continuous at $1$.