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Prove that the sequence ${z_n}$, where $z_n:= x_n-y_n$, converges and $\lim (x_n-y_n)=\lim z_n = (\lim x_n) - (\lim y_n)$

Also, when all limits when $n\to\infty$.

Here is what I have so far: let $x=\lim(x_n), y=\lim(y_n), z=x-y$

let $\epsilon>o$, find an $M_1$ such that for all $n\geq M_1$, we have $$|x_n-x|<\frac{\epsilon}{2};$$find an $M_2$ such that for all $n\geq M_2$, we have $$|y_n-y|<\frac{\epsilon}{2};$$ take $$M:=\max\{M_1, M_2\}$$

For all $n\geq M$ we have:

$$|z_n-z|=|(x_n-y_n)-(x-y)|=|x_n-x-y_n+y|≥ |x_n-x|-|y_n-y|$$

I don't know how to proceed from here, I want to get something like $|z_n-z|<\epsilon$, to complete the proof, right?

Also, is this sufficient enough to prove the second part of the question--the limit part

Please help me here. Thank you!

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Hint:

You're very close! Note that: $$|z_n-z|=|(x_n-y_n)-(x-y)|=|(x_n-x)-(y_n-y)|\leq|x_n-x|+|y_n-y|$$ It was the other side of the triangular inequality!

Please, use MathJax when typing math formulas! :)

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  • $\begingroup$ I see! Thank you so much! :) $\endgroup$ – czhang75 Feb 26 '18 at 19:28
  • $\begingroup$ So far we've proved that it is convergent. Is this sufficient enough to prove the second part of the question--the limit part? $\endgroup$ – czhang75 Feb 27 '18 at 3:12
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    $\begingroup$ You have proved that $z_n\to z=x-y$. So, we have the result. $\endgroup$ – Βασίλης Μάρκος Feb 27 '18 at 13:15
  • $\begingroup$ Awesome! Thanks! $\endgroup$ – czhang75 Feb 27 '18 at 14:42

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