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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 ?} $

4 Answers 4

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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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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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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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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 $b=\sin(x)$.

  • 3
    "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.2011-05-29