$\lim\limits_{} u_n= 0$ and $\lim\limits_{}v_n = 0$ then why $\lim\frac{u_n}{v_n}$ is not defined

Question

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 $r = 0$.

Answer 1

$$\lim\limits_{} u_n= \lim\limits_{} v_n = 0$$ 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.

Answer 2

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.

Answer 3

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.