Question

Find the limit of the sequence
An=(9n+1)/sqrt(25n^2-36)
And
An=n^2/(6^n+8)

Answer 1

\[\lim_{n\to\infty}A_n=\lim_{n\to\infty}\frac{9n+1}{\sqrt{25n^2-36}}=\sqrt{\lim_{n\to\infty}\frac{81n^2+18n+1}{25n^2-36}}\\\ \ \ \ \ \ \ \ \ \ \ =\sqrt{\lim_{n\to\infty}\frac{162n+18}{50n}}=\sqrt{\lim_{n\to\infty}\frac{162}{50}}=\sqrt{\frac{162}{50}}=\sqrt\frac{81}{25}=\frac95\]

Answer 2

Thank you! So the 2nd one doesn't have a limit?

Answer 3

Not sure, I was just too lazy to find it... you can determine that by finding the limit of the ratio between terms and seeing if it's 1 :-)

Answer 4

Lol thanks

Answer 5

yup what he did...