richyw
  • richyw
improper integral. sorry for all of these. my textbook thinks that "converges" is a suitable explanation. \[\int^{\infty}_0 \tan\frac{1}{x}\text{d}x\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
richyw
  • richyw
I need to say if it converges or diverges.
abb0t
  • abb0t
Take the integral as you would from [0, R]
anonymous
  • anonymous
can you change variables to \(u=\frac{1}{x}\) and then flip the limits of integration?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

abb0t
  • abb0t
Then take the limit of the function as R->∞
richyw
  • richyw
is this sufficient to say it converges? \[\lim_{x \to \infty} \tan\frac{1}{x}=0\]
anonymous
  • anonymous
how?
anonymous
  • anonymous
thing thing has an infinite number of singularities
anonymous
  • anonymous
how about the substitution \[x = \tan^{-1}u\]
richyw
  • richyw
oh right.
anonymous
  • anonymous
or \[x=\cot^{-1}(u)\]
anonymous
  • anonymous
the problem isn't just that the path is infinite,
anonymous
  • anonymous
it is undefined infinitely often
klimenkov
  • klimenkov
What about \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) ?
anonymous
  • anonymous
yep!! convergence....
anonymous
  • anonymous
i think there is a mistake here there is no way this thing converges
anonymous
  • anonymous
\[\lim_{x\to0}\tan(1/x)=\infty\]
richyw
  • richyw
oh wait a second, it does diverge.
anonymous
  • anonymous
graph it and see what it looks like, then tell me it converges...
anonymous
  • anonymous
@satellite73 can we say that as \(x\to0\), it is infinite or is it undefined?
richyw
  • richyw
but the limits iof integration are supposed to be from 1 to infty. so ignore the zero
anonymous
  • anonymous
makes no difference, the problem is not the infinite path tangent is undefined infinitely often
klimenkov
  • klimenkov
Use that \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) and immediately say that it diverges.
anonymous
  • anonymous
@richyw it also has infinite "vertical asymptotes" and hence it is discontinuous over the interval and cannot be integrated!
anonymous
  • anonymous
@satellite73 so, you are saying that the function cannot exist to begin with?
richyw
  • richyw
@klimenkov is that just the first term of the taylor expansion at infinity?
klimenkov
  • klimenkov
If you want Taylor series - yes. But from \(\tan x\sim x,\,x\rightarrow0\) implies \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\).
richyw
  • richyw
how am I supposed to know that \(\tan x \sim x,\;x\to 0\) ?
klimenkov
  • klimenkov
Compute\[\lim_{x\rightarrow0}\frac{\tan x}{x}\]
richyw
  • richyw
AWSOME. ok this is what I was struggling with last night.
richyw
  • richyw
why does doing that show that they behave the same?
richyw
  • richyw
that step is what really confuses me!
richyw
  • richyw
same step was used in the second line of this. I don't understand why I do this!
klimenkov
  • klimenkov
Try to check this theorem in your calculus book.
richyw
  • richyw
believe me I have tried. I even emailed my prof asking where the theorem is. does it have a name?
klimenkov
  • klimenkov
Ok. Did you understand why \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) ?
richyw
  • richyw
no, because I don't understand why \(\tan x \sim x,\;x\to 0\)
klimenkov
  • klimenkov
Do you know that \[\lim_{x\rightarrow0}\frac{\sin x}{x}=1\]
richyw
  • richyw
yes
klimenkov
  • klimenkov
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
  • richyw
yes, I can compute the limit.
richyw
  • richyw
but I don't understand its significance
klimenkov
  • klimenkov
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
  • richyw
right. do you know why?
richyw
  • richyw
I mean why the limit thing, not that definition
klimenkov
  • klimenkov
Or \[\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=1\]
richyw
  • richyw
AHHH that's so obvious!
klimenkov
  • klimenkov
It is equivalent definitions.
richyw
  • richyw
omg. thank you so much.
richyw
  • richyw
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
  • richyw
thank you so much!
klimenkov
  • klimenkov
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.