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Let the $\lim_{n\to \infty} x_n=a$ and let $\lim_{n\to \infty} y_n=b$.

Let $\epsilon>0$ for some $n\ge N$

My notes says: $|(x_n)(y_n)-ab|=|x_n(y_n-b)+b(x_n-a)|$

Can someone show me the intermediate steps to the LHS of the equation?

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Just distribute: Notice

$$ x_n(y_n-b) + b(x_n-a) = y_nx_n - x_nb + bx_n - ab = y_nx_n - ab $$

This is just a trick.

In order to show that $x_ny_n \to ab$ provided $x_n \to a$ and $y_n \to b$, we must show by definition that for any $\epsilon > 0$, we can find some $N > 0$ such that if $n > N$, then

$$ |x_ny_n - ab | < \epsilon $$

That is the reason why the starting point is $|x_ny_n - ab|$, because we want to bound this term.

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  • $\begingroup$ I understand the trick. My question was how |(x_n)(y_n)-ab| was arrived at besides the algebraic manipulation. What was the motivation? $\endgroup$ – guest Mar 9 '15 at 3:40
  • $\begingroup$ @guest $|x_ny_n-ab|$ is not arrived at, it is the starting point of the proof. In order to show $x_ny_n\to ab$, you have to show $|x_ny_n-ab|<\epsilon$ for $n\ge N$, so you start with $|x_ny_n-ab|$, then manipulate it until you can show it is less than $\epsilon$. $\endgroup$ – Mike Earnest Mar 9 '15 at 3:47
  • $\begingroup$ Admittedly, "arrived" was a poor and illogical choice of word. What is the motivation for |xnyn−ab|?I want to know why that is the starting point for the argument. $\endgroup$ – guest Mar 9 '15 at 3:49
  • $\begingroup$ guest: see my update. $\endgroup$ – user203867 Mar 9 '15 at 3:51
  • $\begingroup$ So it is just: x_n=a,y_n=b so, (x_n)(y_n)=ab<epsilon [(x_n)(y_n)-ab]<epsilon Then, manipulate LHS . . . $\endgroup$ – guest Mar 9 '15 at 3:54

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