Precalculus (6th Edition) Blitzer

The statement “If $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\ne f\left( a \right)$ and $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists. I can redefine $f\left( a \right)$ to make f continuous at a.” makes sense.
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, $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\ne f\left( a \right)$ and $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists. So the function does not satisfy the third condition of being continuous at a. If the value of $f\left( a \right)$ is redefined and is taken equal to $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$, then, $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ and the function satisfies all the conditions of being continuous Then the function is continuous at a. Hence, the statement makes sense.