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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

I need to say if it converges or diverges.

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

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

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

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

how?

thing thing has an infinite number of singularities

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

oh right.

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

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

it is undefined infinitely often

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

yep!! convergence....

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

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

oh wait a second, it does diverge.

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

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

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

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

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

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

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

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

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

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

why does doing that show that they behave the same?

that step is what really confuses me!

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

Try to check this theorem in your calculus book.

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

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

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

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

yes

yes, I can compute the limit.

but I don't understand its significance

right. do you know why?

I mean why the limit thing, not that definition

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

AHHH that's so obvious!

It is equivalent definitions.

omg. thank you so much.

thank you so much!