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!!!
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: 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}. $