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How do I show that if

$$\lim_{n \to \infty} (x_n-y_n) = 0$$

Then $$\lim_{n \to \infty} x_n \neq \lim_{n \to \infty} y_n$$

Intuitively, this seems false and they should converge to the same limit...

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  • $\begingroup$ I presume you mean $n \to \infty$? $\endgroup$ – copper.hat Nov 25 '13 at 0:58
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    $\begingroup$ If the difference converges to $0$, and either sequence converges, then both converge to the same limit. $\endgroup$ – Daniel Fischer Nov 25 '13 at 0:59
  • $\begingroup$ Yeah, this statement is definitely not true: take $x_n$ and $y_n$ to be the same sequence. $\endgroup$ – NotAwake Nov 25 '13 at 0:59
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    $\begingroup$ This is untrue, if $x_n \to x$ and $x_n-y_n \to 0$, then $y_n \to x$. $\endgroup$ – copper.hat Nov 25 '13 at 0:59
  • $\begingroup$ Triangle inequality should show the limits are equal. $\endgroup$ – user99680 Nov 25 '13 at 1:11
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If $\displaystyle \lim_{n \to +\infty} (x_n-y_n)=0$ then:

  • Both of the sequences are convergente and have the same limit or

  • both of theme are divergents.

Examples for the second case:

1) $x_n$ =n and $y_n=n+ \frac{1}{n+1}$

2) $x_n=\cos n$ and $y_n=\frac{1}{n+1} + \cos n $

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As pointed out in the comments, it can be true that $\lim_{n\to\infty}(x_n + y_n)=0$ and $\lim_{n\to\infty}x_n = \lim_{n\to\infty}y_n$, and so the statement as you've written it in the prompt is false. At the same time, however, it's also false that for any $\{x_n\}$ and $\{y_n\}$ if $\lim_{n\to\infty}(x_n + y_n)=0$ then $\lim_{n\to\infty}x_n = \lim_{n\to\infty}y_n$. A counterexample is $x_n = n$ and $y_n = -n$.

What is true is that if $\lim_{n\to\infty}(x_n + y_n)=0$ and either of the limits $\lim_{n\to\infty}x_n$ and $\lim_{n\to\infty}y_n$ exist, then both limits exist and $\lim_{n\to\infty}x_n = \lim_{n\to\infty}y_n$.

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Here is a proof that the statement is false.

Suppose $\lim_{n \to \infty}(y_n - x_n) = 0$ and suppose $x = \lim x_n$ and $y = \lim y_n$.

I claim that $x=y$. To this end, let $\epsilon > 0$.

Then $|y-x| = |y - y_n + y_n - x_n + x_n - x| \leq |y-y_n| + |y_n - x_n| + |x_n - x|$

Now there exists an $N$ such that for all $n \geq N$ we have that $|y-y_n|, |y_n-x_n|, |y_n - y| \leq \epsilon/3$.

So we have that for any $\epsilon > 0$, $|y-x| \leq \epsilon \Rightarrow y = x$.

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