Formal Definition Of Limit with Sequence

Question

Suppose $\lim_{n\to\infty}x_n = 0$ and the sequence $\{y_n\}_{n=1}^\infty$ is bounded.

Using the formal definition of limit, show that

$$\lim_{n\to\infty}x_ny_n = 0$$

Not sure what is the meaning of the sequence being bounded and how to use the formal definition of limit to show that result.

Answer 1

Replace the sequence $(y_n)$ with a constant sequence $c_n = K, n\in\mathbb N$ where $K>0$ satisfies $\lvert y_n\rvert \leq K, n\in\mathbb N$ (such a $K$ exists, because?). Then it's clear that for every $n\in\mathbb N$ we have $\lvert x_ny_n\rvert\leq \lvert x_nK \rvert$. We know $x_n\to 0$, so for a fixed $\varepsilon >0$, if $n$ is large enough we have $\lvert x_n\rvert < \varepsilon$.

Now, for $\frac{\varepsilon}{K}>0$ there exists $M\in\mathbb N$ such that for every $n\in\mathbb N$ $$n\geq M \implies \lvert x_ny_n\rvert\leq\lvert x_nK\rvert < \frac{\varepsilon}{K}\cdot K = \varepsilon.$$

Answer 2

$\{y_n\}_{n=1}^\infty$ is bounded means: there is $c>0$ such that $|y_n| \le c$ for all $n$.

You have to show: is $ \epsilon >0$, then there is $N \in \mathbb N$ such that $|x_ny_n| <\epsilon $ for all $n > N$.

To this end observe that $|x_ny_n| \le c|x_n|$ and $x_n \to 0$.

Can you proceed ?