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Given this limit $\displaystyle\lim_{n \to{+}\infty}{\frac{\sqrt{16n^2+3}}{(1+a_n)n+5cos n}=\frac{7}{6}}$ I need to calculate this one : $\displaystyle\lim_{n \to{+}\infty}{a_n}$

Any ideas of how to solve it. Thanks!!!

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    Hint: Divide top and bottom by $n$, let $n$ get big. Top approaches $4$. Term $(5\cos n)/n$ at the bottom dies. So for large $n$, $4/[(1+a_n)]$ is very close to $7/6$, and therefore $\dots$2012-05-22

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Hint: write $ {\sqrt{16n^2+3}\over (1+a_n) n +5\cos n} = {n\cdot\sqrt{16+{3\over n^2} }\over n\cdot\bigl( (1+a_n)+{5\cos n\over n}\bigr)} = {\sqrt{16+{3\over n^2} }\over (1+a_n)+{5\cos n\over n}}. $ Then note $ \lim_{n\rightarrow\infty} {\sqrt{16+{3\over n^2} }\over (1+a_n)+{5\cos n\over n}} ={4\over 1+\lim\limits_{n\rightarrow\infty}a_n}. $

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    Perfect. Than$k$ you so much. I have a test tomorrow, and this exercise was ki$l$ling $m$e!2012-05-22