Answer
The statement is true.
Work Step by Step
1) Since $\lim\limits_{x\to a}f(x)$ exists, we know that $\lim\limits_{x\to a^+}f(x)=\lim\limits_{x\to a^-}f(x)=u $
Since $\lim\limits_{x\to a}g(x)$ does not exist, we can say that $\lim\limits_{x\to a^+}g(x)=v $ and $\lim\limits_{x\to a^-}g(x)=w $ and $ v\ne w $
2) Now we consider
$\lim\limits_{x\to a^+}[f(x)+g(x)]=\lim\limits_{x\to a^+}f(x)+\lim\limits_{x\to a^+}g(x)=u+v $
$\lim\limits_{x\to a^-}[f(x)+g(x)]=\lim\limits_{x\to a^-}f(x)+\lim\limits_{x\to a^-}g(x)=u+w $
3) However, since $ v\ne w $, we see that $ u+v\ne u+w $
So, $\lim\limits_{x\to a^+}[f(x)+g(x)]\ne\lim\limits_{x\to a^-}[f(x)+g(x)]$
That means $\lim\limits_{x\to a}[f(x)+g(x)]$ does not exist.
Therefore, the statement is true.
