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

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

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

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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