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Given $$ \begin{cases} \lim_{n\to\infty}x_n = x \ne 0 \\ \lim_{n\to\infty}y_n = \pm\infty \end{cases} $$ Show that for $x < 0$: $$ \lim x_ny_n = \mp\infty $$

I've started with the case when $\lim y_n = +\infty$:

$$ \forall \varepsilon >0\ \exists N_1 \in \Bbb N:\forall n>N_1 \implies y_n > \varepsilon $$

On the other hand the definition for $\lim x_n$ is going to be: $$ \forall \varepsilon > 0 \ \exists N_2 \in \Bbb N: \forall n > N_2 \implies |x_n - x| < \varepsilon $$

That means we may choose some $N$ starting from which the following is true: $$ \forall \varepsilon > 0 \ \exists N = \max\{N_1, N_2\}: \forall n> N \implies y_n > \varepsilon \ \text{and}\ |x_n - x| < \varepsilon $$

So having that in mind we may consider the following system: $$ \begin{cases} |x_n - x| < \varepsilon \\ y_n > \varepsilon \end{cases} \iff \begin{cases} \begin{align} -\varepsilon < &x_n -x < \varepsilon \\ -&y_n < -\varepsilon \end{align} \end{cases} $$

So now if we multiply the inequalities one may obtain: $$ -y_n(x_n - x) < -\varepsilon^2 \iff -x_ny_n + y_nx < \varepsilon^2 \iff x_ny_n>-\varepsilon^2 + y_nx $$

At this point I got stuck, basically my idea was to utilize the definition of limits and then combine the two cases to arrive at a definition of a limit but for the sequence $x_ny_n$, but not sure how to proceed.

What steps should I take to prove what's in the problem section?

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  • $\begingroup$ Is the $\varepsilon-\delta$ way obligatory? $\endgroup$ – Rebellos Dec 9 '18 at 14:09
  • $\begingroup$ @Rebellos not necessarily, that was just my try $\endgroup$ – roman Dec 9 '18 at 14:10
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Suppose $\lim y_n = +\infty$:

We want to show that for any $M>0$, we can find $N>0$, such that if $n>N$, then $x_ny_n < -M$.

We know that

$$ \forall M >0\ \exists N_{1,M} \in \Bbb N:\forall n>N_{1,M} \implies y_n > M $$

In paticular,

$$\exists N_{1,\frac{2M}{|x|}} \in \Bbb N:\forall n>N_{1,\frac{2M}{|x|}} \implies y_n > \frac{2M}{|x|} \iff (-x)y_n > 2M \iff xy_n < -2M $$

On the other hand the definition for $\lim x_n$ is going to be: $$ \forall \varepsilon > 0 \ \exists N_{2,\varepsilon} \in \Bbb N: \forall n > N_{2,\varepsilon} \implies |x_n - x| < \varepsilon $$

In particular, $$ \exists N_{2,-\frac{x}2} \in \Bbb N: \forall n > N_{2,-\frac{x}2} \implies |x_n - x| < -\frac{x}2 \implies x_n < \frac{x}2. $$

Hence for any $n> \max\left(N_{1,\frac{2M}{|x|}} , N_{2,-\frac{x}2}\right), x_ny_n < \frac{x}{2}y_n < \frac{(-2M)}{2}=-M$

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  • $\begingroup$ Could you please elaborate on the parts starting with "In particular, "? $\endgroup$ – roman Dec 9 '18 at 14:31
  • $\begingroup$ I choose my particular $M$ to be $\frac{2M}{|x|}$. and for the seocnd part, I choose my particular $\varepsilon$ to be $-\frac{x}2$ $\endgroup$ – Siong Thye Goh Dec 9 '18 at 14:33
  • $\begingroup$ Oh, i see now, thanks. $\endgroup$ – roman Dec 9 '18 at 14:35

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