## Precalculus (6th Edition) Blitzer

Consider the function $y=\frac{1}{x}$, For $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, Approach the point $x=-1$ on the graph from the left side; the value of the $y-$ coordinate approaches to $-1$. So, $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=-1$, For $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, Now approach the point from the right side. As the point is reached, the value of the $y-$ coordinate approaches to $-1$. So, $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=-1$, Since, $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=-1$, the value for the given limit exists and is given by, $\underset{x\to -1}{\mathop{\lim }}\,\frac{1}{x}=-1$