hey, i have a question that i cannot figure out.

Given that lim(n->infinity)=an exists when a1=1 and an+1=sqrt[1+(6/an)] for n>1, find the limit.

how do i go about solving this?

cheers!

Printable View

- Mar 16th 2009, 05:12 PMjackm7finding the limit
hey, i have a question that i cannot figure out.

Given that lim(n->infinity)=an exists when a1=1 and an+1=sqrt[1+(6/an)] for n>1, find the limit.

how do i go about solving this?

cheers! - Mar 16th 2009, 05:24 PMpankaj
$\displaystyle a_{n+1}=\sqrt{1+\frac{6}{a_n}}$

$\displaystyle \lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty}\sqrt{1+\frac{6}{a_n}}$

$\displaystyle Let \lim_{n\to \infty}a_{n}=a$,then we also have $\displaystyle \lim_{n\to \infty}a_{n+1}=a$

Therefore,$\displaystyle a=\sqrt{1+\frac{6}{a}}$

$\displaystyle a^2=1+\frac{6}{a}$

$\displaystyle a^2-a-6=0$

$\displaystyle

a=3

$ - Mar 16th 2009, 05:34 PMjackm7
thank you, i was getting different answers every time i did that question.

i have also stumbled across another ones i am having problems with.

lim(n->infinity) (1+1/2n)^n-1

lim(n->infinity) ln(n+cos(n))/ln(n^2)

any help would be appreciated - Mar 16th 2009, 05:45 PMpankaj
$\displaystyle

\lim_{n\to \infty}((1+\frac{1}{2n})^{2n})^{\frac{n-1}{2n}}

$

=$\displaystyle

\lim_{n\to \infty}((1+\frac{1}{2n})^{2n})^{\frac{1}{2}-\frac{1}{2n}}

$

=$\displaystyle \sqrt{e}$

Apply L'Hospital for the second limit