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

1 Answer
Euan S.
Jul 27, 2016

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.

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