## Calculus: Early Transcendentals 8th Edition

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.