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

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. – Siminore Jun 10 '12 at 12:23
@Siminore: OK. Thanks. – user 1618033 Jun 10 '12 at 12:24
Numerically, it seems to me that the limit is $-\infty$. – Siminore Jun 10 '12 at 12:36
@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. – user 1618033 Jun 10 '12 at 12:45
@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. – Ragib Zaman Jun 10 '12 at 14:44

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

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

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

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

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sorry, i am new to the site. how do you type symbols here? – tulasi Jun 10 '12 at 14:37
@stone: wow. Very simple! Thanks for your great solution. – user 1618033 Jun 10 '12 at 14:39
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. – Ragib Zaman Jun 10 '12 at 14:39
@st0ne: if i could, i'd give you a golden badge! :-) – user 1618033 Jun 10 '12 at 14:45
@st0ne: how did you think of this solution? – user 1618033 Jun 10 '12 at 14:47