Question

How do you find the limit of `xtan(1/(x-1))` as x approaches infinity?

Answer 1

The limit is 1. Hopefully someone on here can fill in the blanks in my answer.

Explanation:

The only way I can see to solve this is to expand the tangent using a Laurent series at `x=oo`. Unfortunately I've not done much complex analysis yet so I cannot walk you through how exactly that is done but using Wolfram Alpha http://www.wolframalpha.com/input/?i=laurent+series+tan(1%2F(x-1)) I obtained that

`tan(1/(x-1))` expanded at `x =oo` is equal to:

`1/x + 1/x^2 + 4/(3x^3) + 2/(x^4) + 47/(15x^5) + O(((1)/(x))^6)`

Multiplying by the x gives:

`1 + 1/x + 4/(3x^2) + 2/(x^3) + ...`

So, because all the terms apart from the first have an x on the denominator and constant on the numerator

`lim_(xrarroo) (1 + 1/x + 4/(3x^2) + 2/(x^3) + ...) = 1`

because all terms after the first will tend to zero.