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Let $u_n$ be a function $\mathbb{N} \rightarrow \mathbb{R}$ and $v_n$ be a function $\mathbb{N} \rightarrow \mathbb{R}$ such that : $v_n, u_n \ne 0$ for all $n \in \mathbb{N}$ and such that : $\lim\limits_{} u_n= \lim\limits_{} v_n = 0$.

I would like to understand intuitively why we can't say that : $\lim\limits_{} \frac{u_n}{v_n} = 1$. I know many counter-example why this doesn't work, but I don't understand that intuitively. I mean is there a nice way to explain why this is false ?

Why when $\lim\limits_{} u_n= \lim\limits_{} v_n = r$, $r \in \mathbb{R}^{*}$ we have : $\lim\limits_{} u_n/v_n = 1$, I don't see why it makes a difference in the assessment when $r = 0$.

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    $\begingroup$ Why aren't the "many counterexamples" convincing? I mean...take $u_n=\frac 1n$, $v_n=\frac 1{2n}$. The ratio is constant... $\endgroup$ – lulu Jan 16 '18 at 19:31
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    $\begingroup$ Does it not suffice to say that, even if both $u_n$ and $v_n$ converge to $0$, one of them can converge faster than the other one? (Sorry about imprecision here, but we are not putting a mathematical argument anyways but are working on building up intuition...) $\endgroup$ – user491874 Jan 16 '18 at 19:31
  • $\begingroup$ @user8734617 The problem with this is that I can say the same thing for any $r \in \mathbb{R}$ $\endgroup$ – Salutsalut1 Jan 16 '18 at 19:32
  • $\begingroup$ But then we are moving the onus from trying to understand why we cannot say $\lim_{n\to\infty}\frac{u_n}{v_n}=1$ if $r=0$ to trying to understand why we can say $\lim_{n\to\infty}\frac{u_n}{v_n}=1$ if $r\ne 0$. In other words, maybe first you get yourself acquainted with the fact that for $r=0$ you cannot say anything about the limit, and then, in the next step, you find another argument which helps you see that, for $r\ne 0$ there is a limit after all? $\endgroup$ – user491874 Jan 16 '18 at 19:38
  • $\begingroup$ @Salutsalut1 Consider two cases $\,v_n = u_n / 2\,$ and $\,v_n = 2 u_n\,$: what is $\lim u_n / v_n$ in each case? $\endgroup$ – dxiv Jan 16 '18 at 23:44
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$$\lim\limits_{} u_n= \lim\limits_{} v_n = 0$$ tells us intuitively, that when $n$ gets larger and larger $v_n,u_n$ are becoming smaller and smaller.

So, when you look at $\lim_n \frac{u_n}{v_n}$ you are really asking:

What happens if I divide a quantity which becomes smaller and smaller by another quantity which becomes smaller and smaller?

It should not take too long to convince yourself that the correct answer is It depends. It really matters how the two quantities are related.

The question is a (more complicated version of): What do you get if you divide a small number by another small number. Then answer is not 1, it is you could get anything.

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The reason is that $\frac{r}{r}=1$ is true only for $r\neq 0$ thus you can calculate the limit by algebraic rules. When $v_n$ and $u_n$ tend to 0 you need to consider “how” they tend to 0 and in this case the limit can assume any value or do not exist at all.

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For example, let $x_n=\frac{1}{n}$ and $y_n=\frac{1}{n^2}$. Then, $x_n\rightarrow 0$ and $y_n\rightarrow 0$, but $$ \frac{x_n}{y_n} =n\rightarrow \infty.$$

Now, let $x_n=\frac{1}{n^2}$ and $y_n=\frac{1}{n}$. Then, $x_n\rightarrow 0$ and $y_n\rightarrow 0$, but $$ \frac{x_n}{y_n} =\frac{1}{n}\rightarrow 0.$$

Then we can to conclude that in this case the limit is undefined.

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