Answer
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 $.
Work Step by Step
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.
