Answer
The statement “If $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne f\left( a \right)$ and $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( a \right)$, I can redefine $ f\left( a \right)$ to make f continuous at a” does not make sense.
Work Step by Step
For a function to be continuous at a point a, the function must satisfy the following three conditions:
(a) f is defined at a.
(b) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists.
(c) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$
Since, $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne f\left( a \right)$
So, the function does not satisfy the third condition of being continuous at a.
Since $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( a \right)$, so $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ do not exist.
So the function does not satisfy the second condition of being continuous at a.
Redefining $ f\left( a \right)$ can make it equal to $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ but not equal to $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$.
The function does not satisfy all the conditions of being continuous.
Thus, the function is not continuous at a.
Hence, the statement does not make sense.
