# How to solve the limit of a succession on this particular circumstances?

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!!!

-
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 $\dots$ –  André Nicolas May 22 '12 at 22:40

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 $$\lim_{n\rightarrow\infty} {\sqrt{16+{3\over n^2} }\over (1+a_n)+{5\cos n\over n}} ={4\over 1+\lim\limits_{n\rightarrow\infty}a_n}.$$