# 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 $$\lim_{x \to +\infty} a = c\quad\text{and}\quad \lim_{x \to +\infty} b\text{ does not exist ?}$$

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Yes. If the limit of $a+b$ existed, it would follow that

$$\lim_{x \to +\infty}b=\lim_{x \to +\infty} [(a + b) - a]=\lim_{x \to +\infty}(a+b)-\lim_{x \to +\infty}a\;.$$

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Thank you! That helped very much! – crocs May 28 '11 at 23:21

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,

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Thank you! That helped very much! – crocs May 28 '11 at 23:21

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.

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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 $b=\sin(x)$.
when the limit is $\infty$, it does not exist. Writing $\lim\limits_{x\to a}f = \infty$ is a way of saying "the limit does not exist" and specifying why it does not exist. Similarly with limits "equal" to $-\infty$", and with limits as $x\to\infty$ or as $x\to-\infty$. Otherwise, limit theorems need to have a whole bunch of exceptions (e.g., it would no longer be true that if the limits of $f$ and $g$ 'exist', then so does the limit of $f+g$ and the latter is equal to their sum). – Arturo Magidin May 29 '11 at 1:17
@Arturo: One could also argue as follows: In general, it's possible to consider convergence in the extended reals with respect to the order topology; this yields the usual definition of the limits $\pm\infty$. However, in that case not only the limits but also the function values may be $\pm\infty$. Then addition wouldn't be defined, so by using addition the OP was implying that the underlying set was just the reals. Allowing infinite limits but only finite values would be like allowing real limits but only rational values, i.e. considering elements outside the set as potential limits. – joriki May 29 '11 at 1:40
BTW: In most books L'H appears in the form: If bla bla and $\lim\frac{f'(x)}{g'(x)}$ exists then bla bla... Does that cover the case when this limit is $\pm \infty$? Of course I realize that this is not really a good point since most Calculus books are very poorly written, but I am used to make a distinction between limit exists and limit is finite. This might come from the fact that I learned Analysis the old way, sequences first, and there convergence is already finite, so we always use the distinction "convergent" vs. "has a limit". For functions we replaced convergence with finite lim – N. S. May 29 '11 at 1:45
"The limit exists" is relative to the set you consider. You can consider the reals, the extended reals, or even the one-point compactification of the reals with just one point at infinity. The limit of the sequence $(-1)^nn$ exists in the one-point compactification but not in the extended reals. This shows that it's not just a matter of adding "enough" limit points, but also of choosing a topology. Again, there's an analogy to the rationals, which can be extended either to the reals or to the $p$-adic numbers. – joriki May 29 '11 at 1:56