# Limits of "cyclometric" functions

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2004-06-30

Hi.

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,

xarctan(1/x)

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.

Thanks

Hans Lindroth.

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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.

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actually, the limit is 0. well that what my maths software tells me anyway

heres the graph of the function.

also the function tends to 1 as x tends to + or - infinity

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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

Mandrake

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