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Let $x_n$ and $y_n$ be sequences in $\mathbb{R}$ such that $\lim(x_n)\ne 0$ and $\lim(x_ny_n)$ exists. Prove that the $\lim(y_n)$ exists.

Q.matin
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  • You can't prove that $\lim (x_ny_n)$, for instance take $x_n=1$ and $y_n=(-1)^n$, for all $\Bbb N$. – Git Gud Mar 22 '13 at 06:54
  • @GitGud. THen the hypothesis is not satisfied. – Jean-Claude Arbaut Mar 22 '13 at 07:09
  • @arbautjc That's right, but that's not a problem here. The OP says he wants to prove that $\lim (x_ny_n)$ exists. Well that's a trivial consequence of the hypothesis. My counter example is in case the hypothesis is $\lim (x_n)\neq 0$ by itself. – Git Gud Mar 22 '13 at 07:17
  • Well, actually, $\lim(x_n y_n)$ is in the hypotheses, so proving it makes no sense in the first place. Also, $y_n$ is obvliously not a subsequence of $x_n y_n$. Everything but the first statement is simply wrong. – Jean-Claude Arbaut Mar 22 '13 at 09:22

3 Answers3

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Denote $a_n=x_n y_n.$ Then $y_n=\dfrac{a_n}{x_n}.$ If exists $\lim\limits_{n\to\infty}{x_n}\ne{0}$ then also exists $\lim\limits_{n\to\infty}{\dfrac{1}{x_n}}=\dfrac{1}{\lim\limits_{n\to\infty}{x_n}}.$ If, in addition, exists $\lim\limits_{n\to\infty}{a_n}=\lim\limits_{n\to\infty}{x_n y_n},$ then exists
$$\lim\limits_{n\to\infty}{y_n}=\lim\limits_{n\to\infty}{\left(a_n\cdot \dfrac{1}{x_n}\right)}=\lim\limits_{n\to\infty}{a_n}\cdot \lim\limits_{n\to\infty}{\dfrac{1}{x_n}}=\\ =\lim\limits_{n\to\infty}{(x_n y_n)}\cdot \lim\limits_{n\to\infty}{\dfrac{1}{x_n}}=\lim\limits_{n\to\infty}{(x_n y_n)}\cdot\dfrac{1}{\lim\limits_{n\to\infty}{x_n}}.$$

M. Strochyk
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  • I am not following, can you add more detail? – Q.matin Mar 22 '13 at 07:04
  • ... then the $lim(a_n)/lim(x_n)=lim(y_n)$? – Q.matin Mar 22 '13 at 07:07
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    Be aware that $\frac{a_n}{x_n}$ does not necessarily exist for all $n$. It's true for $n>n_0$ for some $n_0$, though, since $\lim(x_n)$ is not null. So the conclusion is still valid. – Jean-Claude Arbaut Mar 22 '13 at 07:11
  • @arbautjc Now I am even more confused :( – Q.matin Mar 22 '13 at 07:16
  • @Q.matin. You may have values of $n$ for which $x_n=0$, then the quotient does not exist. But, since $\lim(x_n)=L \neq 0$, you know that for $\epsilon < {L\over2}$, there is a $n_0$ such that for $n>n_0$, $|x_n-L|<\epsilon$, hence $x_n \neq 0$. If the limit is not null, there may be only finitely many null values of $x_n$. – Jean-Claude Arbaut Mar 22 '13 at 08:28
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    To prove convergence of a sequence, you may look only at values $n>n_0$ (you can change finitely many values of your sequence, that will not change convergence). Thus, you can safely assume $n>n_0$ and $x_n \neq 0$. Then, let $a_n=x_n y_n$, which is convergent. And $x_n$ is also convergent, takes no null values (remember $n>n_0$) and has a non null limit, thus $1/x_n$ is also convergent. The product of two convergent sequences is also convergent, so $a_n x_n = y_n$ is convergent. – Jean-Claude Arbaut Mar 22 '13 at 08:32
  • Sorry, in last step, $\frac{a_n}{x_n}=y_n$ is convergent. – Jean-Claude Arbaut Mar 22 '13 at 08:40
  • Thanks a lot guys! – Q.matin Mar 22 '13 at 21:16
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I give a proof for the case where $y_n$ is bounded.

Suppose $y_n$ is bounded, then we can show that $\forall \epsilon > 0, \exists N_1$ s.t. $\forall n \geq N_1$, $|x_n y_n - x y_n| < \frac{\epsilon}{2} $

Denote $\lim x_n y_n = z$ and $\lim x_n = x$.

Therefore $\forall \epsilon > 0, \exists N_2$ s.t. $\forall n \geq N_2$, $|x_n y_n - z| < \frac{\epsilon}{2} $

We have $|xy_n - z| \leq |xy_n -x_ny_n| + |x_ny_n - z|$.

Therefore for a given $\epsilon > 0$ we have $N = \max (N_1,N_2)$ such that $\forall n \geq N$
$|x y_n - z| < \epsilon$.

Since $x \ne 0$, this proves that $\lim y_n = \frac{z}{x}$.

Sudheer
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Hint: You're told that $\lim (x_n)\ne 0$. What can you do to something which isn't $0$? Use the fact that whenever $\lim (a_n), \lim (b_n)$ exist, then $\lim (a_nb_n)$ exists.

Edit: Using the property that if $\lim (a_n), \lim (b_n)$ exist, then $\lim (a_nb_n)$ exists, let $\displaystyle a_n=\frac{1}{x_n}$ if $x_n\neq 0$ and $a_n=1$ is $x_n=0$, (note that $\lim (x_n)\neq 0$ does not imply that $x_n\neq 0$ for all $n\in \Bbb N$). Set $b_n=x_ny_n$. The limits $\lim (a_n), \lim (b_n)$ exist. You get

$$\lim (a_n)\lim(b_n)=\lim(a_nb_n)=\lim\left(\frac{1}{x_n}\cdot x_ny_n\right)=\lim (y_n)$$

this tells you that $\lim (y_n)=\lim (a_n)\lim(b_n)\in \Bbb R$, which means $\lim (y_n)$ exists.

Git Gud
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  • It's not clear to me from the problem statement that one is allowed to assume $\lim_{n\to\infy}x_n$ exists. If it does exist, it's not allowed to be zero, but does the wording imply it exists? – Gerry Myerson Mar 22 '13 at 06:55
  • @GerryMyerson I'm not sure. I mean, as it is I believe it implies the limit exists, because if it doesn't exist then the formula $\lim (x_n)\neq 0$ doesn't have a truth value as it isn't a statement (it doesn't make sense). But I'm not sure if whoever wrote this problem wants us to assume it exists. Anyway, if it doesn't exist, then $x_n=y_n=(-1)^n$ for all $n\in \Bbb N$ is a counterexample. – Git Gud Mar 22 '13 at 06:58
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    @Q.matin I'm having trouble understanding what your issue is, however I believe that M. Strochyk's answer will help you. – Git Gud Mar 22 '13 at 07:05
  • @Q.matin I added some stuff to the answer which is basically what M. Strochyk wrote in his answer. – Git Gud Mar 22 '13 at 07:22