# Prove $\mathop {\lim }\limits_{x \to \pm \infty } {a \over x} = 0$ [closed]

How do you prove: $\mathop {\lim }\limits_{x \to \pm \infty } {a \over x} = 0$

-

## closed as off-topic by azimut, Thomas, Stefan4024, T. Bongers, Johannes KloosOct 30 '13 at 18:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – azimut, Thomas, Stefan4024, Community, Johannes Kloos
If this question can be reworded to fit the rules in the help center, please edit the question.

What does $\dfrac a{\pm \infty}$ mean to you? – Git Gud Oct 30 '13 at 17:21
I'm afraid you're going to be more accurate: do you mean that the limit of fraction the numerator of which is a constant whereas its denominator is something that diverges to $\;\pm\infty\;$ is zero? If so please do write this correctly, otherwise write, again, what you mean mathematically. – DonAntonio Oct 30 '13 at 17:21
What do you mean by this? Do you mean that you wish to show that $$\lim_{x\to\pm\infty}\frac ax=0$$ or something similar? – Cameron Buie Oct 30 '13 at 17:21
Yes, you both are right. I ment the limit of this expression – captain dragon Oct 30 '13 at 17:26

Unfortunately, that isn't enough to prove it by contradiction--you'll only be showing that the limit, if it exists, isn't $\epsilon$.

It would be better to proceed directly. By definition, $\lim\limits_{x\to\pm\infty}\dfrac{a}{x}=0$ if and only if for every $\epsilon>0$ there is some $M\ge 0$ such that $\left|\dfrac{a}{x}-0\right|<\epsilon$ whenever $|x|>M$.

Well, with a little rewriting, note that $$\left|\frac{a}{x}-0\right|=\left|\frac{a}{x}\right|=\frac{|a|}{|x|},$$ so for $x\ne 0$ and $\epsilon>0,$ we have that $\left|\dfrac{a}{x}-0\right|<\epsilon$ is equivalent to $\dfrac{|a|}{\epsilon}<|x|.$ (Why?) What can we do now?

-
isn't it enough to say $|a| / \epsilon$ must be less than infinity? – captain dragon Oct 30 '13 at 17:53
or maybe show that $|a|/\epsilon$ isn't bounded? – captain dragon Oct 30 '13 at 17:55
No, it is not. Regardless of your choice of $a$ and regardless of your choice of $\epsilon>0,$ we will always have that $\frac{|a|}\epsilon$ is a real number. Moreover, if $a\ne 0,$ then this will not be bounded as we allow $\epsilon$ to vary. Take a look again at the definition. We picked an arbitrary $\epsilon,$ and we're looking for an $M$ such that...what? How does that rewrite help us, then? – Cameron Buie Oct 30 '13 at 17:55
ok, so our M should be for example $M = \left[ {{{|a|} \over \varepsilon }} \right] + 1$. – captain dragon Oct 30 '13 at 18:19
Yes, that will work, though there's no real need for $M$ to be an integer. If we put $M=\frac{|a|}{\epsilon}+1,$ Then if $|x|\ge M,$ we will have $|x|\ne 0$ and $\frac{|a|}{\epsilon}<|x|,$ so that $\left|\frac{a}{x}-0\right|<\epsilon,$ as desired. Apologies for the delayed reply. – Cameron Buie Nov 4 '13 at 18:26