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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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I suggest that the account of @crocs be suspended. I don't think there's any point in further efforts at cooperation; there is clearly no intent whatsoever of cooperating on his or her part. I don't usually object to strong language, but if the only response to diverse respectful and patient attempts at explaining the problems is just calling two of the people undertaking those efforts a bitch, that's a bit too much. –  joriki May 29 '11 at 2:52
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@Bill: It is not the OP's prerogative to "emphasize what they desire in whatever manner they so desire" - profanity is unacceptable, for example. Would you really allow, say, "this f---ing problem is really a piece of s---, can someone help me"? At any rate, even if the OP's post is within the bounds of decency, as crocs' was, as the FAQ indicates, "If you are not comfortable with the idea of your questions and answers being edited by other trusted users, this may not be the site for you." –  Zev Chonoles May 29 '11 at 19:39
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Please take this conversation to meta. (Bill, I don't know why you're automatically siding with the OP, whose offensive comments were all deleted by the time I got here, when there is still clear evidence that many other reasonable users were responding to very offensive behavior on the OP's part. I think you enjoy the narrative of the rest of math.SE oppressing new users too much to ask yourself whether it's actually true in this case.) –  Qiaochu Yuan May 29 '11 at 20:21
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@Bill: Know better than to... what, exactly? Make minor, yet justified, edits? I don't understand why you think experienced users should have to walk on eggshells on the off-chance the OP might start hurling profanity as soon as something happens that they don't like. Nothing anyone else did could possibly have been expected to provoke crocs' reaction - it was completely out of proportion. There isn't anything anyone could have done to avoid it; this was simply a consequence of crocs having a bad temper. –  Zev Chonoles May 29 '11 at 20:45
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@Bill: You do not know whether this could have easily been avoided because you do not know the facts of the case. The users who were present before all of the offensive comments were deleted actually know the facts, and you should be deferring to their judgment. Otherwise you have nothing to fall back on but your own prejudices against the system. Your comment that "users memories of heated conversations are not always reliable" is ridiculous: you don't even have any memories of the relevant incident, so why is your judgment more trustworthy than theirs? –  Qiaochu Yuan May 29 '11 at 20:46

4 Answers 4

up vote 25 down vote accepted

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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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 $b=\sin(x)$.

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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
    
I think that this depends from book to book and instructor to instructor. And the sum theorem, holds as long as it is not an indeterminate form, I am also pretty sure I saw somewhere that theorem stated for finite limits.. –  N. S. May 29 '11 at 1:37
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@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
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"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

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