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I would like to prove that if we're in the context of a limit for $x \to \infty$ of a function $f$ (lets suppose that $f$ has finite limit $l$) we can suppose $x>0$ and, more in general, we can suppose $x>a$ with $a>0$.

My approach is that we know by limit definition that for all $\varepsilon>0$ exists $K_\varepsilon>0$ such that for all $x>K_{\varepsilon}$ it is $|f(x)-l|<\varepsilon$.

So from this the first statement is immediate, since by $x>K_{\varepsilon}>0$ for transitivity it is $x>0$; so, for the other statement, since the definition of limit works for $K_{\varepsilon}$ it works for all $M_{\varepsilon}>K_{\varepsilon}$, so for proving that we can suppose $x>a$ it is enough to use the arbitrarity of $M_{\varepsilon}$. Lets choose $M_{\varepsilon}$ such that $x-a>M_{\varepsilon}-a>K_{\varepsilon}>0$, hence it is $x>M_{\varepsilon}>a$.

Is this correct? If not, can someone point out my mistakes? Thanks.

I suppose that this works for $x\to x_0$ too, in the sense that if $x \to x_0$ we can suppose $|x-x_0| < a$ for all $a>0$; indeed for $x \to x_0$ the definition of limit works for a $\delta_{\varepsilon}>0$ such that $|x-x_0|<\delta_{\varepsilon}$ then it works for all $\sigma_{\varepsilon}<\delta_{\varepsilon}$, and so it is enough to take $\sigma_{\varepsilon}>a$ so we have $x_0-\delta_{\varepsilon}<x_0-\sigma_{\varepsilon}<x_0-a<x<x_0+a<x_0+\sigma_{\varepsilon}<x_0+\delta_{\varepsilon}\Rightarrow |x-x_0|<a$.

Has this sense?

A colleague of mine wanted to prove the first statement (the one of limit as $x\to\infty$) with the theorem (which I know has the name "permanence of sign theorem") that if $f \to l$ with $l>0$ for $x\to c$ then $f(x)>l$ for $|x-c|<\delta_{\varepsilon}$, but it looks wrong to me because it gives us information on the sign of $l$ and not on the sign of $x$; is my colleague wrong or right? Thanks.

ZaWarudo
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  • You question is pretty unclear. Do you want to say that $\lim _{x \rightarrow \infty} f(x)=L \implies x>0$? Because that is what the title suggests and it does not make so much sense. – Eminem Jun 03 '20 at 15:22
  • @eminem: Thanks for your answer, sorry if my question looks unclear. I want to find a rigorous way for the concept "for big $x$", used in in the context of limit; that concept, sometimes, is used to suppose $x>0$ or $x>a$ for the independent variable $x$. But where it comes from? And the argument I've written is a try to make it rigorous, but I'm not sure if it is correct. An analogous is when $x \to x_0$ that is used for suppose $x-x_0<a$ for $a>0$. – ZaWarudo Jun 03 '20 at 15:28
  • I'm still not sure what you question is. There are (very) rigor and formal definitions to all the kinds of limits you can imagine and a quick search will find them. Could you be more specific in what you are looking for? – Eminem Jun 03 '20 at 15:30
  • @eminem: I'll try to clarify with an example. Suppose you want to evaluate the limit as $x \to \infty$ of $\sqrt{x^2-1}-x$, one way is writing it as $\sqrt{x^2(1-1/x^2)}-x=|x|\sqrt{1-1/x^2}-x$. Now, to get rid of the modulus, you can suppose that $x>0$ because $x \to \infty$ and it is perfectly clear to me in a intuitive way, but how that supposition justified with rigour? That's what I've tried to prove in the first case in my question, and in general I tried to prove it for other cases such limit as $x\to \infty$ of $\sqrt{(x-1)^2-1}-x$ supposing $x>1$ (in general $x>a$). Hope is clear now! – ZaWarudo Jun 03 '20 at 15:43
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    So you question is that if $x \rightarrow \infty \implies x>0$? Because if it is, then yes. What you formally want is to prove $\infty > a \forall a> 0$ but that is a different area. If you realy want to prove this you will need Extended Real Numbers. Take a look here – Eminem Jun 03 '20 at 15:52
  • @eminem: Something like that, exactly; and more in general that implies $x>a$ for all $a\in\mathbb{R}$. Thanks for the link that I will read as soon as I can! However I think that it isn't needed, and that's why I tried to prove it with the limit definition. – ZaWarudo Jun 03 '20 at 15:59

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Well, the definition of $\lim_{x \to \infty}=\infty$ is that for every $M>0$ there exists $x_0>0$ such that for every $x>x_0$ $f(x)>M$.

So either I don't understand the question or that the answer is as above.

Orenio
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