## Precalculus (6th Edition) Blitzer

The complete statement is, “If $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=L$ and $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=M$ where $L\ne M$, then $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ does not exist."
Consider the provided limit notation, $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=L$ Here, $f$ is any function defined on some open interval containing the number $a$. The function f may or may not be defined at a. Hence, the notation $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=L$ means that as $x$ gets closer to $a$ from the left, but remains unequal to $a$, the corresponding value of $f\left( x \right)$ gets closer to L. Consider the other provided limit notation, $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=M$; this implies that as $x$ gets closer to $a$ from the right, but remains unequal to $a$, the corresponding value of $f\left( x \right)$ gets closer to M. Since, $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=L$ and $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=M$ there is no single number that the values of $f\left( x \right)$ are close to when x is close to a. Hence we can conclude, “If $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=L$ and $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=M$ where $L\ne M$, then $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ does not exist.