## Precalculus (6th Edition) Blitzer

The limit exists for the function $f\left( x \right)$ at $x=5$ and is equal to $\underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)=\underline{0}$.
To find the limit of the given curve, first observe the graph near $x=5$ from both sides. As $x$ approaches $5$ from the left, the values of $f\left( x \right)$ get closer to the $y-$ coordinate of the point shown by the closed dot $\left( 5,0 \right)$. The $y-$ coordinate of this point is 0 and the left-hand limit is 0. Thus, as $x$ gets closer to $5$ from left, the values of $f\left( x \right)$ get closer to 0. Therefore, $\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0$ As $x$ approaches $5$ from the right, the values of $f\left( x \right)$ get closer to the $y-$ coordinate of the point $\left( 5,0 \right)$. Thus, as $x$ gets closer to $2$ from right, the values of $f\left( x \right)$ get closer to $5$. Therefore, $\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f\left( x \right)=0$ Because, $\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0\text{ and }\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f\left( x \right)=0$, $f\left( x \right)$ is close to $0$ when $x$ is close to $5$. Therefore, as $x$ approaches 5, $\underset{x\to 5}{\mathop{\lim }}\,f\left( x \right)=0$