# Given $\lim_{n\to\infty}x_n = x \ne 0$ and $\lim_{n\to\infty}y_n = \pm\infty$ show that $\lim x_ny_n = \mp\infty$ when $x<0$

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?

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

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

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