dunkelheit
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- Sep 7, 2018
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- 48
To give some context: I was trying to evaluate
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.
[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.
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