#### Answer

The statement “f and g are both continuous at $ a $, although $ f+g $ is not” 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 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.