$\lim (a + b)\;$ when $\;\lim(b)\;$ does not exist? Suppose $a$ and $b$ are functions of $x$. Is it guaranteed that
$$
\lim_{x \to +\infty} a + b\text{ does not exist}
$$
when
$$
\lim_{x \to +\infty} a = c\quad\text{and}\quad
\lim_{x \to +\infty} b\text{ does not exist ?}
$$
 A: Just to complete joriki's answer:
His answer is right only if $c$ is a number, which seems to be the case from the way you wrote the question. 
Anyhow if you deal also with infinite limits,  it is possible that $\lim_{x \to \infty}b$ does not exist and $\lim_{x \to \infty} a+b$ exists.
Just take $a=x-\sin(x)$ and $b=\sin(x)$.
A: Yes. If the limit of $a+b$ existed, it would follow that
$$\lim_{x \to +\infty}b=\lim_{x \to +\infty} [(a + b) - a]=\lim_{x \to +\infty}(a+b)-\lim_{x \to +\infty}a\;.$$
A: HINT $\ $ This follows immediately from the fact that functions whose limit exists at $\rm\:\infty\:$ are closed under subtraction, i.e. they comprise an additive subgroup of all functions. Therefore, abstractly, this is essentially the same as the proof that the sum of an integer and non-integer is a non-integer. For further discussion of this complementary view of a subgroup see this post. 
A: Suppose, to get a contradiction, that our limit exists.  That is, suppose $$\lim_{x\rightarrow \infty} a(x)+b(x)=d$$ exists.  Then since $$\lim_{x\rightarrow \infty} -a(x)=-c,$$ and as limits are additive, we conclude that $$\lim_{x\rightarrow \infty} a(x)+b(x)-a(x)=d-c$$ which means $$\lim_{x\rightarrow \infty} b(x)=d-c.$$  But this is impossible since we had that $b(x)$ did not tend to a limit.
Hope that helps,
