Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!!!

share|improve this question
    
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
add comment

1 Answer

up vote 1 down vote accepted

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

share|improve this answer
    
Perfect. Thank you so much. I have a test tomorrow, and this exercise was killing me! –  limoragni May 22 '12 at 23:12
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.