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)=M $ where $ L\ne M $, then $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ does not exist."
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)=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.
