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Suppose that $\left \{ x_n \right \}$, $\left \{ y_n \right \}$ are sequences with $|x_n-x_m|\leq \frac{1}{n}$ for all $n\in\mathbb{N}$, and suppose that $\lim_{n\rightarrow \infty }{x_n} = L$. Prove that $\lim_{n\rightarrow \infty }{y_n} = L$.

My attempt: Since $\lim_{n\rightarrow \infty }{x_n} = L$, we know that for all $\epsilon>0$ there exists an $N$ so that for any $n>N$, $|x_n-L|<\epsilon$.

So for any $n>N$, $|y_n-L| = \left| y_n - x_n + x_n - L \right| \Rightarrow \leq \left| y_n - x_n \right|+\left| x_n - L\right|= \left| x_n - y_n \right| + \left| x_n - L \right|\leq \frac{1}{n}+\epsilon$.

Where do I go from here?

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    $\begingroup$ In your first sentence, do you mean $|x_n - y_n| \le 1/n$? $\endgroup$ – Christopher A. Wong Dec 3 '13 at 21:07
  • $\begingroup$ I don't think that "$\Rightarrow$" belongs there. $\endgroup$ – Barry Cipra Dec 3 '13 at 21:18
  • $\begingroup$ Looks like you mean $|x_n-y_n| \le \frac1{n}$. Otherwise, you have no hypotheses about $y_n$. $\endgroup$ – marty cohen Dec 3 '13 at 22:08
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You've nearly done it. You found that $|y_n - L| \le 1/n + \epsilon$ for all $n > N$. Since for any choice of $\epsilon$, for sufficiently large $n$ we have $1/n < \epsilon$, we can substitute that into the upper bound you found. Can you complete it from here?

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  • $\begingroup$ If u can please show me the substitution step then I think I can finish it off. $\endgroup$ – user87274 Dec 3 '13 at 22:16
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    $\begingroup$ So you get that for each $\epsilon > 0$, we have that for all sufficiently large $n$, $|y_n - L| \le \epsilon + \epsilon$. Compare this with the definition for the limit of a sequence. $\endgroup$ – Christopher A. Wong Dec 4 '13 at 0:36

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