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I need to find the limit according to alpha. I appreciate any hint enter image description here

Anna
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    As it is, the question makes no sense: What is tending to what?? Besides this, read the following so that you can actually write the question properly and not that pic: http://meta.matheducators.stackexchange.com/questions/93/mathjax-basic-tutorial-and-quick-reference – Timbuc Nov 17 '14 at 14:58

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$$\begin{align}\lim_{n\to\infty}\frac{(n+1)^2+\alpha(n-1)^2}{(n-2)^2+(n+2)^2} &=\lim_{n\to\infty}\frac{(\alpha+1)n^2+n(1-\alpha)+(1+\alpha)}{2n^2+4}\\&= \lim_{n\to\infty}\frac{(\alpha+1)+\frac{(1-\alpha)}{n}+\frac{(1+\alpha)}{n^2}}{2+\frac{4}{n^2}}\\ &=\frac{\alpha+1}{2}\\ \end{align}$$

Aditya Hase
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Hint. As $n$ tends to $+\infty$, $$ a_n \sim \frac{(1+\alpha)\:n^2}{2\:n^2}$$ The limit is easy to find out now :)

Euler's student
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