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Compute the following limit:

$ \lim_{n\to\infty} \frac{1}{n}\int_0^n \frac{\arctan(x)}{\arctan{\frac{n}{x^2-nx+1}}}dx$

I'm looking for an easy approach if possible.

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    @Siminore, stone resolved the question, it was $1/2$ ! I was misled by numerical data as well, mathematica gave me totally bogus values that make me think the limit was $-\infty$ as well.2012-06-10

1 Answers 1

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$\arctan \left(\frac{n}{x^{2}-nx+1}\right) = \arctan(x) + \arctan(n-x)$

$ I= \int_{0}^{n} {\frac {\arctan(x)}{ \arctan(x)+\arctan(n-x)}dx} =\int_{0}^{n} {\frac {\arctan(n-x)}{ \arctan(x)+\arctan(n-x)}dx} $

$I = \frac{1}{2}\cdot\int_{0}^{n} 1dx = \frac{n}2$

$\lim_{n \to \infty }\frac1n \int_{0}^{n} \frac {\arctan(x)}{ \arctan \left(\frac{n}{x^{2}-nx+1}\right)}dx = \frac12 $

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    how do we get the step $I = \frac{1}{2}\cdot\int_{0}^{n} 1dx$2012-06-10