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richyw
 3 years ago
improper integral. sorry for all of these. my textbook thinks that "converges" is a suitable explanation.
\[\int^{\infty}_0 \tan\frac{1}{x}\text{d}x\]
richyw
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0I need to say if it converges or diverges.

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

anonymous
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0Then take the limit of the function as R>∞

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thing thing has an infinite number of singularities

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the problem isn't just that the path is infinite,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it is undefined infinitely often

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep!! convergence....

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

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

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

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

anonymous
 3 years 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
 3 years 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

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

anonymous
 3 years 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!

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

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

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

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

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

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

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

richyw
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2Try to check this theorem in your calculus book.

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

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

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

klimenkov
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0yes, I can compute the limit.

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

klimenkov
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0I mean why the limit thing, not that definition

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

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

richyw
 3 years 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
 3 years 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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