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Suppose $\lim\limits_{n\to\infty} y_n=L$ and $|x_n-y_n|\le\frac1n$ for all $n$. Show that $\lim\limits_{n\to\infty} x_n=L$.

Hi! Can anyone help me with this real analysis sequence problem. I'm not really sure what to do with the following: $$|x_n-y_n| \leq \frac1n .$$

Any help is appreciated! Thanks!

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closed as off-topic by Simply Beautiful Art, Shailesh, Namaste, Jack, Leucippus Sep 14 '17 at 1:15

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  • $\begingroup$ Hint: add zero. $\endgroup$ – Sean Roberson Sep 13 '17 at 21:53
  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. To help get you started, I retyped the text from the picture. $\endgroup$ – Martin Sleziak Sep 14 '17 at 0:20
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    $\begingroup$ Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) $\endgroup$ – Martin Sleziak Sep 14 '17 at 1:40
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$$|x_n - L | =|x_n - y_n + y_n - L | \leq |x_n - y_n | + |y_n - L| $$

Now let $n\to \infty$.

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We have $$\lim_{n\rightarrow\infty} (x_n-y_n)=0$$

Therefore , we have $$\lim_{n\rightarrow\infty} x_n=\lim_{n\rightarrow\infty} (x_n-y_n)+\lim_{n\rightarrow\infty}y_n=0+L=L$$

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For any $\epsilon > 0$, there exists an $N$ such that when $n>N$

$|y_n - L| < \frac{\epsilon}{2}$ and $|y_n-x_n| \le \frac {1}{n} < \frac{\epsilon}{2}$

Then:

$|x_n - L| = |(y_n-L) - (y_n-x_n)| <|y_n-L| + |y_n-x_n| < \frac {\epsilon}{2} + \frac {\epsilon}{2}$

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