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

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- schrodinger

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- richyw

I need to say if it converges or diverges.

- abb0t

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

- anonymous

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

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## More answers

- abb0t

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

- richyw

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

- anonymous

how?

- anonymous

thing thing has an infinite number of singularities

- anonymous

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

- richyw

oh right.

- anonymous

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

- anonymous

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

- anonymous

it is undefined infinitely often

- klimenkov

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

- anonymous

yep!! convergence....

- anonymous

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

- anonymous

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

- richyw

oh wait a second, it does diverge.

- anonymous

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

- anonymous

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

- richyw

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

- anonymous

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

- klimenkov

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

- anonymous

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

- anonymous

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

- richyw

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

- klimenkov

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

- richyw

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

- klimenkov

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

- richyw

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

- richyw

why does doing that show that they behave the same?

- richyw

that step is what really confuses me!

- richyw

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

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- klimenkov

Try to check this theorem in your calculus book.

- richyw

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

- klimenkov

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

- richyw

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

- klimenkov

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

- richyw

yes

- 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

yes, I can compute the limit.

- richyw

but I don't understand its significance

- 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

right. do you know why?

- richyw

I mean why the limit thing, not that definition

- klimenkov

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

- richyw

AHHH that's so obvious!

- klimenkov

It is equivalent definitions.

- richyw

omg. thank you so much.

- 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

thank you so much!

- 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.

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