## Precalculus (6th Edition) Blitzer

The complete statement is, “If $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=L$ and $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=L$ then $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=L$.
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, $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=L$; this means 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 L. Since, both the limit notation $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=L$ and $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=L$ are the same, this implies that the value of $f\left( x \right)$ at $x=a$ from the left and from the right is the same. Thus, if $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=L$ and $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=L$ then $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=$ L.