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

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 $x = \infty$. 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

$\tan \left(\frac{1}{x - 1}\right)$ expanded at $x = \infty$ is equal to:

$\frac{1}{x} + \frac{1}{x} ^ 2 + \frac{4}{3 {x}^{3}} + \frac{2}{{x}^{4}} + \frac{47}{15 {x}^{5}} + O \left({\left(\frac{1}{x}\right)}^{6}\right)$

Multiplying by the x gives:

$1 + \frac{1}{x} + \frac{4}{3 {x}^{2}} + \frac{2}{{x}^{3}} + \ldots$

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

${\lim}_{x \rightarrow \infty} \left(1 + \frac{1}{x} + \frac{4}{3 {x}^{2}} + \frac{2}{{x}^{3}} + \ldots\right) = 1$

because all terms after the first will tend to zero.