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Let $(x_n)_{n \in \mathbb{N}}$ and $(y_n)_{n \in \mathbb{N}}$ be sequences in $\mathbb{R}$ such that $x_n \to \pm \infty$ and $(x_ny_n)_{n \in \mathbb{N}}$ converges. Show that $y_n \to 0$.

This problem is driving me crazy. All I did so far was to state the following definitions:

  • If $x_n \to \infty$, then for all $M > 0$, there exists an $n_1 \in \mathbb{N}$ such that $n \ge n_1$ implies $x_n > M$.

  • If $x_n \to -\infty$, then for all $N < 0$, there exists an $n_2 \in \mathbb{N}$ such that $n \ge n_2$ implies $x_n < N$.

  • If $x_ny_n \to L$, where $L \in \mathbb{R}$, then for all $\epsilon > 0$, there exists an $n_3 \in \mathbb{N}$ such that $n \ge n_3$ implies $|x_n y_n-L|<\epsilon$.

How can I approach this problem? I thought about proving by contradiction, that is, what happens if $L > 0$ and if $L < 0$, for each case of $x_n \to +\infty$ and $x_n \to -\infty$?

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Since $(x_n y_n)$ converges to say $L$ so let $\epsilon>0$ then $\exists n_0$ and for $n\ge n_0$ we have

$$M|y_n-\frac L{x_n}|\le |x_n||y_n-\frac L{x_n}|=|x_ny_n-L|<\epsilon$$ hence by the triangle inequality we get $$0\le|y_n|<\frac{\epsilon}M+\frac L{|x_n|}<\frac{2\epsilon}M,\;\text{for $n$ large}$$

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  • $\begingroup$ I get that $|x_n| > M$ means $\frac 1{|x_n|} < \frac 1M$, but how do we have $L < \epsilon$ in the last line? $\endgroup$ – Cookie Oct 29 '14 at 0:13
  • $\begingroup$ The sequence $\left(\frac L{|x_n|}\right)$ converges to $0$ since $x_n\to+\infty$ hence for $n$ large enough we have $$\frac L{|x_n|}<\frac{\epsilon}{M}$$ $\endgroup$ – user63181 Oct 29 '14 at 0:16
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You have arrived at

$$|x_n y_n-L|<\epsilon\Leftrightarrow L-\epsilon < x_ny_n < L+\epsilon, $$ from where (if $x_n\ne 0$)

$$\frac{L-\epsilon}{|x_n|} < |y_n| < \frac{L+\epsilon}{|x_n|} .$$ In particular,

$$|y_n| < \frac{L+\epsilon}{|x_n|} .$$

Now, since $|x_n|\to \infty,$ for any $N>0$ there exists $n_0$ such that $n\ge n_0 \implies |x_n|>N.$ Thus, for $n\ge n_0$ it is

$$ |y_n| < \frac{L+\epsilon}{|x_n|}<\frac{L+\epsilon}{N} .$$

Can you finish now?

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Note that a convergent sequence is bounded. Thus: For some $M > 0$, and $m \in \mathbb{N}$, we have:

$|x_ny_n| < M$ for $n \geq m$, thus: $|y_n| < \dfrac{M}{|x_n|}$, this implies $y_n \rightarrow 0$.

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hint: you only need the fact that $x_ny_n$ is bounded. Use the composition theorem with the function $ x\to \frac 1x $

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Let $(x_n)$ and $(a_n)$ be sequences of real numbers such that $x_n\to\pm\infty$ and $a_n\to L$ for some $L\in\mathbb{R}$.

Then $\displaystyle\frac{a_n}{x_n}\to 0$ since $\displaystyle\frac{a_n}{x_n}=\big(a_n\big)\left(\frac{1}{x_n}\right)$ where $\big(a_n\big)$ is bounded and $\displaystyle\frac{1}{x_n}\to 0$, so

applying this result with $a_n=x_n y_n$ gives $y_n\to 0$.

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