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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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    I have just improved your TeX syntax, since the double fraction was really too small.2012-06-10
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    @Siminore: OK. Thanks.2012-06-10
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    Numerically, it seems to me that the limit is $-\infty$.2012-06-10
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    @Siminore: really? i tried to use W|A but i failed. Probably i need W|A Pro version in order to get some more time.2012-06-10
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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

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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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    sorry, i am new to the site. how do you type symbols here?2012-06-10
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    @stone: wow. Very simple! Thanks for your great solution.2012-06-10
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    Welcome to math.stackexchange! Excellent answer! Chris and I were struggling over it for a while over in a chat room. To see what I've changed about your post to render your LaTeX, or to see what anyone has typed to make any post, you can just click "edit" on their post and see.2012-06-10
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    @st0ne: if i could, i'd give you a golden badge! :-)2012-06-10
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    @st0ne: how did you think of this solution?2012-06-10
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    how do we get the step $I = \frac{1}{2}\cdot\int_{0}^{n} 1dx$2012-06-10