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Let $\{x_n\}$ and $\{y_n\}$ be convergent sequences. Prove that the sequence $\{z_n\},$ where $z_n=x_n y_n,$ converges and $\lim_{n\to\infty} (x_n y_n) =\lim_{n\to\infty} z_n = (\lim_{n\to\infty} x_n)(\lim_{n\to\infty} y_n).$

(Partial proof) Let $\varepsilon>0$ be given. As ${x_{n}}$ is convergent, it is bounded. Therefore, we find a $B>0$ s.t. $|x_{n}|\le B $ for all $n\in N$. Next We find $M_{1}$ s.t. for all $n\ge M_{1} $ we have $$|x_{n}-x|\le \frac{\varepsilon}{2(|y|+1)} $$

We find $M_{2}$ s.t. for all $n\ge M_{2} $ we have $$|y_{n}-y|\le \frac{\varepsilon}{2B}. $$

I didn't write the whole proof. I would like to ask this, why do we use $\dfrac{\varepsilon}{2(|y|+1)}$ and $\dfrac{\varepsilon}{2B}$? or how do we get them?

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The goal is to make $|x_ny_n-xy|$ smaller than $\epsilon,$ and the trick is to use the triangle inequality: $$ |x_ny_n-xy|\leq |y_n-y||x_n|+|x_n-x||y|.$$ Since $|x_n|\leq B$ for all $n,$ they make $|y_n-y|<\frac{\epsilon}{2B}$ so that $|y_n-y||x_n|<\frac{\epsilon}{2},$ and $|x_n-x|<\frac{\epsilon}{2(|y|+1)}$ so that, since $\frac{|y|}{|y|+1}<1,$ we have $|x_n-x||y|<\frac{\epsilon}{2}.$ Then $$ |x_ny_n-xy|\leq |y_n-y||x_n|+|x_n-x||y|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon,$$ as desired.

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  • $\begingroup$ Now I could clearly see how we get there, we make it smaller in order to obtain both $|x_{n}-x|$ and $|y_{n}-y|$ with using triangle inequality. $\endgroup$ – user460584 Jul 17 '17 at 23:27

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