Precalculus (6th Edition) Blitzer

The statement “f and g are both continuous at $a$, although $f+g$ is not” does not make 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 f and g are continuous functions at a, both of them satisfy the above three conditions. Now check whether the function $f+g$ is continuous at a, or not. Find the value of $\left( f+g \right)\left( x \right)$ at $a$, $\left( f+g \right)\left( a \right)=f\left( a \right)+g\left( a \right)$ Since, $f\left( a \right)$ and $g\left( a \right)$ are defined, so is $\left( f+g \right)\left( a \right)$. Now find the value of $\,\underset{x\to a}{\mathop{\lim }}\,\left( f+g \right)\left( x \right)$, $\,\underset{x\to a}{\mathop{\lim }}\,\left( f+g \right)\left( x \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$. Since, $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ and $\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ both exist, thus $\,\underset{x\to a}{\mathop{\lim }}\,\left( f+g \right)\left( x \right)$ also exists. From the above steps, $\,\underset{x\to a}{\mathop{\lim }}\,\left( f+g \right)\left( x \right)=f\left( a \right)+g\left( a \right)=\left( f+g \right)\left( a \right)$ Thus, the function satisfies all the properties of being continuous. Thus, the function $f+g$ is continuous at a. Hence, the statement does not make sense.