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 2 years ago
quick question: there is no meaning in the concept of approaching infinity from the right, correct?
\(x\to\infty^+\)
 2 years ago
quick question: there is no meaning in the concept of approaching infinity from the right, correct? \(x\to\infty^+\)

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TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2or \(x\to\infty^\) wither, right?

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0When the variable is x, and it takes on only positive values, then it becomes positively infinite. We write \[x\to\infty^+\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0you gotta start really really far out i guess

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2@mathslover no, I think you are thinking of \(x\to+\infty\)

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0I'm still trying to figure out this: \[\int\limits \frac{x \dot\ \sin x}{1 + \cos^2x} dx\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2I mean, infinity is not a number even how do you approach a nonnumber from a higher number no number can be higher than a nonnumber seems like asking if 7>purple

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2@Hero it has been solved, and I will be happy to give you hints should you desire let me note that that integral is unsolvalble (at least by me) as an indefinite integral, you must put the bounds

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1\(∞^\pm\) aren't even in the Extended real number line

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0I want to solve it without the bounds. I already saw the solution with bounds. Since the function is continuous I will assume an indefinite integral exists.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2Good luck with that! I don't know how to solve it without making use of the interval 0 to pi that it is supposed to be

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0Apparently no one or no thing does. Not even mathematica, geogebra, maple or TI. But not to worry, I know one person who for sure will be able to tell if it is possible. If it is integrable this guy will know. I kinda miss @JamesJ

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2me too :) @zarkon actually taught me how to solve this, but he made use of the bounds heavily

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0I contacted TI about this today and they wanted me to provide them with the solution to this so that they could use it to update their algorithm.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2wow, far out I remember one thinbg was that we proved that\[\int_0^\infty{x\sin x\over1+\cos^2x}dx\]does not have a finite value, though I forget how we proved it. that suggests to me that no closed form of the indefinite integral exists, but I could be wrong

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0Hmmm interesting. It may not have a finite value, but that doesn't mean it can't be solved. Last time I checked, if you take the integral of something and get sin(x) as the result....well sin(x) isn't "finite" either, but it still exists as a solution

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0There are many indefinite integrals with solutions that are not finite.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2true... I am stretching my capacities here admittedly

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0Well, if by finite you mean a specific value.
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