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richyw
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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
 one year ago
richyw Group Title
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
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

electrokid Group TitleBest 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!
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

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

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

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

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

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

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

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

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

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

richyw Group TitleBest ResponseYou've already chosen the best response.0
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!
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
thank you so much!
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago
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