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 one year ago
improper integral. sorry for all of these. my textbook thinks that "converges" is a suitable explanation.
\[\int^{\infty}_0 \tan\frac{1}{x}\text{d}x\]
 one year ago
improper integral. sorry for all of these. my textbook thinks that "converges" is a suitable explanation. \[\int^{\infty}_0 \tan\frac{1}{x}\text{d}x\]

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richyw
 one year ago
Best ResponseYou've already chosen the best response.0I need to say if it converges or diverges.

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Take the integral as you would from [0, R]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0can you change variables to \(u=\frac{1}{x}\) and then flip the limits of integration?

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Then take the limit of the function as R>∞

richyw
 one year ago
Best ResponseYou've already chosen the best response.0is this sufficient to say it converges? \[\lim_{x \to \infty} \tan\frac{1}{x}=0\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0thing thing has an infinite number of singularities

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0how about the substitution \[x = \tan^{1}u\]

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0or \[x=\cot^{1}(u)\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0the problem isn't just that the path is infinite,

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0it is undefined infinitely often

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2What about \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) ?

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0yep!! convergence....

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0i think there is a mistake here there is no way this thing converges

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x\to0}\tan(1/x)=\infty\]

richyw
 one year ago
Best ResponseYou've already chosen the best response.0oh wait a second, it does diverge.

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0graph it and see what it looks like, then tell me it converges...

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0@satellite73 can we say that as \(x\to0\), it is infinite or is it undefined?

richyw
 one year ago
Best ResponseYou've already chosen the best response.0but the limits iof integration are supposed to be from 1 to infty. so ignore the zero

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0makes no difference, the problem is not the infinite path tangent is undefined infinitely often

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2Use that \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) and immediately say that it diverges.

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0@richyw it also has infinite "vertical asymptotes" and hence it is discontinuous over the interval and cannot be integrated!

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0@satellite73 so, you are saying that the function cannot exist to begin with?

richyw
 one year ago
Best ResponseYou've already chosen the best response.0@klimenkov is that just the first term of the taylor expansion at infinity?

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2If you want Taylor series  yes. But from \(\tan x\sim x,\,x\rightarrow0\) implies \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\).

richyw
 one year ago
Best ResponseYou've already chosen the best response.0how am I supposed to know that \(\tan x \sim x,\;x\to 0\) ?

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2Compute\[\lim_{x\rightarrow0}\frac{\tan x}{x}\]

richyw
 one year ago
Best ResponseYou've already chosen the best response.0AWSOME. ok this is what I was struggling with last night.

richyw
 one year ago
Best ResponseYou've already chosen the best response.0why does doing that show that they behave the same?

richyw
 one year ago
Best ResponseYou've already chosen the best response.0that step is what really confuses me!

richyw
 one year ago
Best ResponseYou've already chosen the best response.0same step was used in the second line of this. I don't understand why I do this!

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2Try to check this theorem in your calculus book.

richyw
 one year ago
Best ResponseYou've already chosen the best response.0believe me I have tried. I even emailed my prof asking where the theorem is. does it have a name?

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2Ok. Did you understand why \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) ?

richyw
 one year ago
Best ResponseYou've already chosen the best response.0no, because I don't understand why \(\tan x \sim x,\;x\to 0\)

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2Do you know that \[\lim_{x\rightarrow0}\frac{\sin x}{x}=1\]

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2Now \[\lim_{x\rightarrow0}\frac{\tan x}{x}=\lim_{x\rightarrow0}\frac{\frac{\sin x}{\cos x}}{x}=\lim_{x\rightarrow0}\frac{\sin x}{x}=1\]because \(\lim_{x\rightarrow0}\cos x=1\). Got it?

richyw
 one year ago
Best ResponseYou've already chosen the best response.0yes, I can compute the limit.

richyw
 one year ago
Best ResponseYou've already chosen the best response.0but I don't understand its significance

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2Ah. Ok. If \(f(x)\sim g(x),\,x\rightarrow a\), it means that \(f(x)\) behaves like \(g(x)\) at the point \(x=a\).

richyw
 one year ago
Best ResponseYou've already chosen the best response.0right. do you know why?

richyw
 one year ago
Best ResponseYou've already chosen the best response.0I mean why the limit thing, not that definition

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2Or \[\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=1\]

richyw
 one year ago
Best ResponseYou've already chosen the best response.0AHHH that's so obvious!

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2It is equivalent definitions.

richyw
 one year ago
Best ResponseYou've already chosen the best response.0omg. thank you so much.

richyw
 one year ago
Best ResponseYou've already chosen the best response.0I feel so stupid now, but if you looked at my questions history I spent like 3 hours last night trying to get this explanation!

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2Don't be shy to ask. It is normal not to see obvious things because obvious is a relative concept. I wish you a lot of luck in learning math. Love it.
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