# Real Analysis sequences help please [closed]

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!

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

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$$|x_n - L | =|x_n - y_n + y_n - L | \leq |x_n - y_n | + |y_n - L|$$

Now let $n\to \infty$.

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

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

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:

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