Positivity of a variable when a limit tends to infinity

[dunkelheit]

To give some context: I was trying to evaluate

[MATH]\lim_{x \to \infty} \int_x^{2x} e^{-t^2} \text{d}t[/MATH]​

I know that this limit can be easily evaluated using De L'Hôpital, but I'm trying to tackle it with estimations and because of that I had a doubt. I notice that if [MATH]t \geq x \geq 0[/MATH] it is [MATH]e^{-t^2} \leq e^{-x^2}[/MATH] and so

[MATH]0 \leq \lim_{x \to \infty} \int_x^{2x} e^{-t^2} \text{d}t \leq \lim_{x \to \infty} \int_x^{2x} e^{-x^2} \text{d}t= \lim_{x \to \infty} e^{-x^2} =0[/MATH]​

Hence by comparison the limit is [MATH]0[/MATH]: the doubt is that [MATH]t \mapsto e^{-t^2}[/MATH] is decreasing if [MATH]t>0[/MATH], so I need that [MATH]t>0[/MATH] to make that estimation.
Intuitively I see that since [MATH]x \to \infty[/MATH] the interval of integration [MATH][x,2x][/MATH] will be, sooner or later, an interval that contains positive values of [MATH]x[/MATH] and so I can suppose that [MATH]x>0[/MATH], but I don't know formally why is like that. I suppose that it comes somehow from the definition of limit as [MATH]x \to \infty[/MATH].

Can someone show me from where I can suppose that [MATH]x>0[/MATH] if I have to deal with a limit as [MATH]x \to \infty[/MATH]? Thanks.

 

[Cubist]

I would write, "Consider the case when x>0, which is clearly true in the limit as \(x \to \infty \)"

Whenever anyone writes "clearly" that's the bit to check very closely . But I personally think you're on safe ground here.

BTW: I think you made a slight error...

[MATH] \lim_{x \to \infty} \int_x^{2x} e^{-x^2} \text{d}t[/math]
[math]= \lim_{x \to \infty}\left( e^{-x^2} \int_x^{2x}1\text{d}t \right)[/MATH]
[math]= \lim_{x \to \infty}\left( xe^{-x^2} \right) [/MATH]
[math]=0 [/math]
EDIT: Perhaps that last step needs a few extra lines now? I used a computer algebra system to verify that the limit was 0

 

[dunkelheit]

Thanks for your answer, actually the [MATH]x[/MATH] vanishes because I've made a typo; the limit was

[MATH]\lim_{x \to \infty} \frac{1}{x} \int_x^{2x} e^{-t^2} \text{d}t[/MATH]​

This explain why I forgotten to write that [MATH]x[/MATH] (and I've made it more than one times because of copy-paste ), it actually vanishes with [MATH]\frac{1}{x}[/MATH].
However I still don't get it completely. The only thing that makes sense to me in this behaviour of assuming [MATH]x>0[/MATH] when [MATH]x \to \infty[/MATH] (apart of the intuitively reason) is that we're in a context of limit, so it is satisfied that for all [MATH]\varepsilon>0[/MATH] there esists [MATH]K_{\varepsilon}>0[/MATH] such that if [MATH]x>K_{\varepsilon}[/MATH] then [MATH]\left|\frac{1}{x} \int_x^{2x} e^{-t^2} \text{d}t-L\right|<\varepsilon[/MATH].

So since [MATH]x>K_{\varepsilon}>0[/MATH] I can assume [MATH]x>0[/MATH], because I'm interested in what happens when this definition is satisfied (i.e. when I take the limit) and when this definition is satisfied it is true that [MATH]x>K_{\varepsilon}[/MATH] which implies that [MATH]x>0[/MATH].
(I've used the definition of a finite limit [MATH]L[/MATH], but the reasoning seems valid too if [MATH]L=\pm \infty[/MATH]).

I hope it is clear and I hope this is correct, what you think? Thanks!

 

[Cubist]

I think you might be over-complicating things. I'm not sure why you need to introduce a new variable \( K_{\varepsilon} \). In this Wikipedia page on limits they seem to use symbol δ in this type of context. And doesn't δ imply a small distance from the target (in this case ∞)? There are others on this forum who know more about this than me, so I hope one of them will respond to you.

I personally think it is sufficient to partition the domain of the original function by stating that you're considering the case x>0. This allows x to approach +∞ and lets you make the simplification.