So I had to explain to someone about limits and I got asked: how do you prove that $\lim\limits_{x\to\infty}\frac1x=0$ without intuition.

After a while I answered:

Let's construct a sequence $(a_n)$ where $a_n=\frac1n$(the sequence starts from $n=1$) now, $\left[\frac1x\right]'=-\frac1{x^2}$ thus $a_n>a_{n+1}$.

Now let's assume that there is a value, say $k$, such that $a_n>k>0,\forall n\in\Bbb Z^+$.

If $k$ is a rational number then $k=\frac qp\ge\frac1p=a_p>a_{p+1}$ which is contradiction

If $k$ is irrational then: $k=\cdots +b_2+b_1+b_0+b_{-1}+b_{-2}+\cdots$ where $b_i=c\times 10^i, c\in\{0,1,2,3,4,5,6,7,8,9\}$. Now I can construct a number $k'=\max\{\,b_i\ne0\}$. $k'$ is rational, hence $k>k'=\frac qp\ge\frac1p=a_p>a_{p+1}$ which is contradiction.

This implies that there is no $k$ such that $a_n>k>0,\forall n\in\Bbb Z^+$ which implies that $\inf\{a_n\}=0$ and because $a_n>a_{n+1}$ I also know that $\inf\{a_n\}=\lim\limits_{x\to\infty}\frac1x$ hence $\lim\limits_{x\to\infty}\frac1x=0$.

My question is, is my proof okay? I don't think that I have any problems but I am not completely sure. And if yes. Is there a easier way to prove that for any positive irrational number there is always smaller positive rational number?

  • $\begingroup$ Btw, I'm well aware about math.stackexchange.com/questions/1205010/… $\endgroup$ – Holo Nov 27 '17 at 15:22
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    $\begingroup$ Do you really mean "without intuition" ? I would have expected the converse. Without intuition, you apply the formal definition of a limit. $\endgroup$ – Yves Daoust Nov 27 '17 at 15:22
  • $\begingroup$ @YvesDaoust no, without intuition means a formal proof, the reason I have this question and not just using the way of the formal definition is because the one that asked me this is not at that level yet $\endgroup$ – Holo Nov 27 '17 at 15:27
  • $\begingroup$ @Holo This seems quite a cute proof. Did you come up with it, on your own? $\endgroup$ – bat_of_doom Nov 27 '17 at 15:29
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    $\begingroup$ Sorry but in my opinion, your proof is much harder than that with the mere definition of a limit, and it invokes more advanced concepts, such as a derivative ! $\endgroup$ – Yves Daoust Nov 27 '17 at 15:34

You want to prove that there exists no $k>0$ such that $$a_n>k>0 \; \forall n \in \mathbb N$$

So, you can choose $$n=\left\lfloor \frac k2 \right \rfloor +1$$

Since, $$n = \left\lfloor \frac k2 \right \rfloor +1 > \frac k2 \implies a_n = \frac{1}{n}< \frac k2 <k $$

This leads to contradiction, and hence we're done.


$$\forall\ \epsilon>0:\forall\ x>\frac1\epsilon:\left|\frac1x-0\right|<\epsilon.$$

This proves constructively that

$$\forall\ \epsilon>0:\exists X:\forall\ x>X:\left|\frac1x-0\right|<\epsilon.$$


you can prove directly by the definition, if you fix $$M>0$$ you can always find a value $\epsilon >0$ $$x=0+\epsilon$$ such that $$\frac {1}{x} > M$$ infact it is sufficient to select $$\epsilon < \frac {1}{M}$$

  • $\begingroup$ The problem is that the one asked me that is not at a level he learned the formal definition of limit $\endgroup$ – Holo Nov 27 '17 at 15:27
  • $\begingroup$ ok you can explain in this way as it was a game: tell your friend to fix a number as big as he can, then you choose $\epsilon$ such that $1/\epsilon$ is grater than M, with the good choice you always win, then the limit is $+\infty$ $\endgroup$ – gimusi Nov 27 '17 at 15:31
  • $\begingroup$ @gimusi you are using the Archimedean property without justifying it, because note that M is an integer. $\endgroup$ – bat_of_doom Nov 27 '17 at 15:36
  • $\begingroup$ @bat_of_doom sorry I can't get your point about the Archimedean property $\endgroup$ – gimusi Nov 27 '17 at 15:39


$\lim_{x \rightarrow \infty} \dfrac{1}{x} = 0.$

Let $\epsilon >0$ be given.

Choose $M > 1/\epsilon$ , $M,$ real.

Then :

For $x > M:$

$|\dfrac {1}{x}| < 1/M < \epsilon.$


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