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Does there exist two sequences $(x_n)_n$, $(y_n)_n$ of real numbers such that $\lim_n x_n-y_n\neq 0$ (may not exist), but $\lim_n x_n+y_n=0$ and $\lim_n x_ny_n=1$?

Notice that it cannot be the case that both sequences are convergent, or even bounded, since by taking convergent subsequences, say with limits $x$ and $y$, we would have that $x+y=0$ and $xy=1$, which has no real solutions.

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2 Answers 2

up vote 23 down vote accepted

Use that

$$(x_n+y_n)^2-4x_ny_n=(x_n-y_n)^2.$$

So $(x_n-y_n)^2$ has a limit. If we take limits then

$$-4=0-4=\lim_{n\to\infty}(x_n+y_n)^2-4\lim_{n\to \infty}x_ny_n=\lim_{n\to\infty}((x_n+y_n)^2-4x_ny_n)=\lim_{n\to\infty}(x_n-y_n)^2.$$

But this gives a contradiction, since $(x_n-y_n)^2\ge 0,\forall n\in \Bbb{N}.$

So, the sequences do not exist.

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Assume toward a contradiction that there are such sequences $(x_n)_n$ and $(y_n)_n$. Because $\lim_n x_n y_n$ is positive, $x_n$ and $y_n$ eventually have the same sign. Without loss of generality, we may assume that for infinitely many $n$ they are both positive. By passing to a subsequence we may assume that $x_n$ and $y_n$ are both positive for all $n$. Then $x_n$ and $y_n$ are both between $0$ and $x_n + y_n$, which approaches zero as $n \to \infty$, so $x_n$ and $y_n$ themselves approach zero as $n \to \infty$. Therefore $\lim_n x_ny_n = 0$, which is a contradiction.

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