Precalculus (6th Edition) Blitzer

The correct answers to fill the blanks provided in the statement “We find $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ using $f\left( x \right)=$_______ and we find $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ using $f\left( x \right)=$_______” are ${{x}^{2}}+5$ and ${{x}^{3}}+1$ respectively.
Consider the piecewise function defined by: f\left( x \right)=\left\{ \begin{align} & {{x}^{2}}+5\text{ if }x<2 \\ & {{x}^{3}}+1\text{ if }x\ge 2 \\ \end{align} \right. To find $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, look at values of $f\left( x \right)$ when $x$ is close to $2$ but less than $2$. Because $x$ is less than $2$, use the first line of the piecewise function’s equation, $f\left( x \right)={{x}^{2}}+5$ where $x<2$. Similarly, to find $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, look at values of $f\left( x \right)$ when $x$ is close to $2$ but greater than $2$. Because $x$ is greater than $2$, use the second line of the piecewise function’s equation, $f\left( x \right)={{x}^{3}}+1$ where $x\ge 2$. Therefore, the complete statement with filled blanks would be “We find $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ using $f\left( x \right)=$${{x}^{2}}+5 and we find \underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right) using f\left( x \right)=$${{x}^{3}}+1$”.