If $(x_n)$ and $(y_n)$ are positive real sequences such that $(y_n)$ is bounded and $\lim\frac{(x_n)}{(y_n)} = 0$, prove that $\lim(x_n) = 0$

So since $(y_n)$ is bounded, there exists a real number $M$ such that for all n natural numbers, $(y_n) < |M|$ So by the limit definition :
$\lim|\frac{(x_n)}{(y_n)}-0| = \lim\frac{(x_n)}{(y_n)} < \epsilon$ So $|(x_n)| < \epsilon(y_n)$

Now I need to find a natural number $N$ such that for all $n > N$, $|(x_n)| < \epsilon$ is this correct which would prove this statement. However I'm having trouble picking something that will work here? Am I on the right track? I have to be very careful also about anything I use to quote the theorem since this is an introduction to real analysis class.



2 Answers 2


I would approach this by contradiction:

If the sequence does not converge to 0, then there is a subsequence that is bounded away from 0. So we may fix $a>0$ such that $|x_n|>a$ for infinitely many values of $n$.

Since the sequence of $y_n$ is bounded, say by $M$, we have $|y_n|<M$ for all $n$, so if $n$ is one of the indices in our subsequence, we have $|x_n/y_n|>a/M$.

This inequality holds for infinitely many $n$, and we have found a subsequence of $(x_n/y_n)$ that does not converge to 0.

Does this make sense?

  • $\begingroup$ Does this take into account the case where lim(x_n) does not exist if you are doing it by contradiction? $\endgroup$
    – Justin
    Nov 30, 2010 at 4:25
  • $\begingroup$ The argument above tells you that the assumption that "it is not the case that $\lim_n x_n=0$" leads to a contradiction. This includes the case where the limit does note exist. $\endgroup$ Nov 30, 2010 at 4:28
  • $\begingroup$ This actually makes a lot of sense and is very easy to understand. It's not the most intuitive way to approach it (from my point of view since I'm just starting with this), so I may not accept it, but I will definitely vote up. EDIT: The only step that I may not be able to justify using the tools I currently have is when you multiply $(x_n)$ and $a$ intothe inequality $\endgroup$
    – Justin
    Nov 30, 2010 at 4:33
  • $\begingroup$ It's not $|x_n|$ that's greater than $a$, it's $|x_{n_{k}}|$ for some k natural number since this is a subsequence? $\endgroup$
    – Justin
    Dec 1, 2010 at 0:29
  • $\begingroup$ I'm just not seeing how a subsequence is relating to every case for $\frac{|x_n|}{|y_n|}$ since we need to show that $\frac{|x_n|}{|y_n|}$ > something for ALL n? $\endgroup$
    – Justin
    Dec 1, 2010 at 1:24


If $|y_{n}|\leq M$ then $1 \leq \frac{M}{|y_{n}|} $

$| x_{n}| \leq |x_{n}| \cdot \frac{M}{|y_{n}|} = |\frac{x_{n}}{y_{n}}| \cdot M < \varepsilon_{0} \cdot M = \frac{\varepsilon}{M} \cdot M = \varepsilon$

Where $|\frac{x_{n}}{y_{n}}| < \varepsilon_{0}$ *for hypotesis* and choosing $\varepsilon_{0} = \frac{\varepsilon}{M}$

  • $\begingroup$ How do you choose $\epsilon_0$? I thought it has to be arbitrary, i.e. any real number $> 0$? $\endgroup$
    – Justin
    Nov 30, 2010 at 3:46
  • $\begingroup$ As $\frac{x_{n}}{y_{n}} \longrightarrow 0$ then $\forall \varepsilon_{0}, \exists N \in \mathbb{N}/ |\frac{x_{n}}{y_{n}}-0|< \varepsilon_{0}$. In particular, you can choose $\varepsilon_{0} =\frac{\varepsilon}{M}$ $\endgroup$ Nov 30, 2010 at 3:51
  • $\begingroup$ You start with $\varepsilon$ arbitrary, but $\varepsilon_{0}$ is geting from $\frac{x_{n}}{y_{n}} \longrightarrow 0$. $\endgroup$ Nov 30, 2010 at 4:01
  • $\begingroup$ I'm a little confused. Do you mean to use the $\epsilon$ variable here? What do you mean "Where $|\frac{x_{n}}{y_{n}}| < \varepsilon_{0}$ for hypotesis? We know that $|\frac{x_{n}}{y_{n}}| < $some varying epsilon but not a fixed epsilon? $\endgroup$
    – Justin
    Nov 30, 2010 at 4:30
  • $\begingroup$ @Fdart17: The definition of limit is ''For every positive number, whatever letter we might denote it by (often but not always $\epsilon$), there exists $N$ such that blabla''. In order to prove something regarding the sequence $(x_n)$, you first fix some arbitrary $\epsilon>0$. Then, to find $N$ such that the limit definition is satisfied for that sequence, you apply the definition of limit to the other sequence $(x_n/y_n)$ not with that number $\epsilon$, but with the positive number $\epsilon_0=\epsilon/M$ instead. $\endgroup$ Nov 30, 2010 at 10:23

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