richyw
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
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I need to say if it converges or diverges.

abb0t
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Take the integral as you would from [0, R]

anonymous
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can you change variables to \(u=\frac{1}{x}\) and then flip the limits of integration?

abb0t
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Then take the limit of the function as R>∞

richyw
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is this sufficient to say it converges? \[\lim_{x \to \infty} \tan\frac{1}{x}=0\]

anonymous
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how?

anonymous
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thing thing has an infinite number of singularities

electrokid
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how about the substitution
\[x = \tan^{1}u\]

richyw
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oh right.

electrokid
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or \[x=\cot^{1}(u)\]

anonymous
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the problem isn't just that the path is infinite,

anonymous
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it is undefined infinitely often

klimenkov
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What about \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) ?

electrokid
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yep!! convergence....

anonymous
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i think there is a mistake here
there is no way this thing converges

electrokid
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\[\lim_{x\to0}\tan(1/x)=\infty\]

richyw
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oh wait a second, it does diverge.

anonymous
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graph it and see what it looks like, then tell me it converges...

electrokid
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@satellite73 can we say that as \(x\to0\), it is infinite or is it undefined?

richyw
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but the limits iof integration are supposed to be from 1 to infty. so ignore the zero

anonymous
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makes no difference, the problem is not the infinite path
tangent is undefined infinitely often

klimenkov
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Use that \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) and immediately say that it diverges.

electrokid
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@richyw it also has infinite "vertical asymptotes" and hence it is discontinuous over the interval and cannot be integrated!

electrokid
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@satellite73 so, you are saying that the function cannot exist to begin with?

richyw
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@klimenkov
is that just the first term of the taylor expansion at infinity?

klimenkov
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If you want Taylor series  yes. But from \(\tan x\sim x,\,x\rightarrow0\) implies \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\).

richyw
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how am I supposed to know that \(\tan x \sim x,\;x\to 0\) ?

klimenkov
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Compute\[\lim_{x\rightarrow0}\frac{\tan x}{x}\]

richyw
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AWSOME. ok this is what I was struggling with last night.

richyw
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why does doing that show that they behave the same?

richyw
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that step is what really confuses me!

richyw
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same step was used in the second line of this. I don't understand why I do this!

klimenkov
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Try to check this theorem in your calculus book.

richyw
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believe me I have tried. I even emailed my prof asking where the theorem is. does it have a name?

klimenkov
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Ok. Did you understand why \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) ?

richyw
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no, because I don't understand why \(\tan x \sim x,\;x\to 0\)

klimenkov
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Do you know that \[\lim_{x\rightarrow0}\frac{\sin x}{x}=1\]

richyw
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yes

klimenkov
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Now \[\lim_{x\rightarrow0}\frac{\tan x}{x}=\lim_{x\rightarrow0}\frac{\frac{\sin x}{\cos x}}{x}=\lim_{x\rightarrow0}\frac{\sin x}{x}=1\]because \(\lim_{x\rightarrow0}\cos x=1\). Got it?

richyw
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yes, I can compute the limit.

richyw
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but I don't understand its significance

klimenkov
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Ah. 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
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right. do you know why?

richyw
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I mean why the limit thing, not that definition

klimenkov
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Or \[\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=1\]

richyw
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AHHH that's so obvious!

klimenkov
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It is equivalent definitions.

richyw
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omg. thank you so much.

richyw
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I feel so stupid now, but if you looked at my questions history I spent like 3 hours last night trying to get this explanation!

richyw
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thank you so much!

klimenkov
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Don'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.