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This might be a very dumb question, but I have googled around and looked in asymptotic analysis reference texts and cannot seem to find what I am looking for. So here we go:

Suppose I know that there exists some $x$ such that $xn \sim n^2$ as $n \to \infty$. Can I conclude that $x \sim n$?

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  • $\begingroup$ Welcome to Maths SX! I suppose $x$ is an expression depending on $n$? $\endgroup$ – Bernard Jul 4 '18 at 16:58
  • $\begingroup$ There are various equivalence relations that you might have in mind. In any case a proof will proceed (perhaps indirectly) from the definition of that equivalence, so stating it in the Question is certainly expeditious. $\endgroup$ – hardmath Jul 4 '18 at 16:59
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Yes, because equivalence is compatible with multiplication, so $$xn\sim n^2\quad\text{and}\quad \frac1n\sim\frac1n\quad\text{imply}\quad xn\cdot \frac1n=x\sim n^2\frac1n=n.$$

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  • $\begingroup$ Ah, thank you. Is there a similar thing I can do for an equation like $x^2 - nx \sim n^2$ — again with $n$ arbitrary large and $x = x(n)$? $\endgroup$ – H. Löw Jul 4 '18 at 17:11
  • $\begingroup$ I'm afraid not (I don't know the context), as equivalence is not compatible with addition nor subtraction. $\endgroup$ – Bernard Jul 4 '18 at 17:27
  • $\begingroup$ For that you would use the method of dominant balance @H.Löw $\endgroup$ – Antonio Vargas Jul 5 '18 at 0:19

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