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We know that $$\lim\limits_{n\to\infty}y_nx_n=\lim\limits_{n\to\infty}y_n.\lim\limits_{n\to\infty}x_n$$ provided two limits exist. Now, Suppose a sequence $\{x_n\}\subseteq\mathbb{R}$ is has the property that $x_n\geq 1$ for all $n\in\mathbb{N}$ and $\{y_n\}\subseteq\mathbb{R}$ is another sequence such that $\lim\limits_{n\to\infty}y_n=\alpha$ for some $\alpha\in (0,\infty]$. $$\text{ Deoes } \lim\limits_{n\to\infty}y_nx_n\geq\alpha?$$

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    Does the limit exist for $x_n$? I mean is it given convergent or divergent or none of these two? – 2017-02-11
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    The limit doesn't necessarily exist, but every limit point of $y_nx_n$ must be $≥\alpha$. – 2017-02-11
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    limit of $x_n$ may and may not exist. What we know is that, all terms of it are greater than or equal to one. – 2017-02-11
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    Let $x_n=\{1, 2, 1, 2, 1, 2...\}$. The product limit can't exist so you cannot use your limit expression. – 2017-02-11

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The equation $$\lim_n x_ny_n = \left( \lim_n x_n \right) \left( \lim_n y_n \right) \qquad (\star)$$ will not hold in general.

If we choose $x_n=2+(-1)^n$ (so we have $x_n\geq 1$ for all $n\in \mathbf N$), then $(x_n)_n$ is not convergent since $\liminf_n x_n \neq \limsup_n x_n$ and therefore $\lim_n x_n$ does not exist. Therefore $(\star)$ does not hold.

If we add that $(x_n)_n$ is monoton decreasing, then $(x_n)_n$ is convergent since it is bounded below and hence the equation $(\star)$ holds.

The inequality $$\lim_n x_ny_n \geq \alpha$$ where $\alpha:= \lim_n y_n$ is always valid since we have $$ \lim_n x_ny_n \geq \lim_n y_n = \alpha $$ where we used $x_n \geq 1$.

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    Thanks, but can I say that $\lim\limits_{n\to\infty}\inf x_n y_n\geq1$. I guess no, also. For example the sequence {1,2,3, ... }. – 2017-02-11
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    Not, but you can say it is bigger than $\alpha$ since $\lim_n x_ny_n \geq \lim_n y_n = \alpha$. I will edit my answer! – 2017-02-11
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If the limit $\lim_{n} (x_n y_n) $ exist then the limit $\leq \alpha $ because..if $\lim_{n} (x_n y_n) $ exist,then as $\lim_{n} (x_n) $ exist and $!=0$ $\Therefore \lim x_n$ exists and as $x_n \leq 1$ $$\Therefore \lim x_n$$ and hence$$ \lim_{n} (x_n y_n) $$ $$=\lim_{n} x_n \lim_{n} y_n \geq \lim_{n} y_n = \alpha $$