How to show that two converging sequences do not have the same limit? 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...
 A: 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 $
A: 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 $\lim_{n\to\infty}x_n = \lim_{n\to\infty}y_n$.
A: 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 $\epsilon > 0$,  $|y-x| \leq \epsilon \Rightarrow y = x$.
