Jump to content

Limits of "cyclometric" functions


Recommended Posts





My name is Hans Lindroth. I study maths at the university for the first time and was hoping someone could answer me this. If I'd like to find the limit of, say,



lim x->0


How would I do it? I mean, does the function arctan(1/x) go to infinity as (1/x) does (when x->0) or does it just oscillate like arcsine or arccos? If anybody knows, please tell me.




Hans Lindroth.

Link to comment
Share on other sites

Okay this problem is relativity easy...first find the Maclaurin Series for arctan and since it is going to 0 you can use the series in the interaval [-1,1]

the maculaurin series is x-x^3/3+x^5/5-x^7/7...now subsititue the series in for arctan u get x( 1/x-1/3x^3+1/x5^5...)

distrubte 1-1/3x^2+1/5x^4... and then plug in zero for the limit and you get undefined...which means it oscillates like arcsin and arccos.

A way to think about it is when ever you have 1/x it means that you basically flip the coordinate grid all the values that go to zero instead goto infinity and vice versa so the question is basically asking what will arctan be at infinity which is unknowable or undefined.

Link to comment
Share on other sites

Bishnu, the problem with your reasoning is that you are using the taylor developpement around zero and then you fill in 1/x, but what you basically would need is the developpement at infinity.

Which is 1/2 Pi - 1/x + 1/3 1/x^3 - 1/5 1/x^5 + ....

filling in 1/x, and multiplying with x gives you an expression with only x's, so taking the limit gives 0



Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.