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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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  • $\begingroup$ 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 $\dots$ $\endgroup$ May 22, 2012 at 22:40

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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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  • $\begingroup$ Perfect. Thank you so much. I have a test tomorrow, and this exercise was killing me! $\endgroup$
    – limoragni
    May 22, 2012 at 23:12

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