# Three similar limit problems

• May 18th 2008, 09:55 AM
simplysparklers
Three similar limit problems
Hey guys!
I have 3 similar problems, if someone could please show me how to do one of them that would be great!
They all involve finding $\lim_{n\rightarrow\infty}$ of $a_{n}$, when $a_{n}$ isn't given.

1.) $a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3}$ $(n\geq{1})$
Find $\lim_{n\rightarrow\infty}a_{n}$

2.) $a_{1}=2, a_{n+1}=\frac{2a_{n}+3}{a_{n}+2}$ $(n\geq{1})$
Find $\lim_{n\rightarrow\infty}a_{n}$

3.) Determine the limit of $(a_{n})$, where
$a_{1}=2,a_{n+1}=\sqrt{2+2a_{n}}$ $(n\geq{1})$

If someone could show me how to do one of them I'd really appreciate it!!

Thanks,
Jo
• May 18th 2008, 10:01 AM
Moo
Hello,

Quote:

Originally Posted by simplysparklers
Hey guys!
I have 3 similar problems, if someone could please show me how to do one of them that would be great!
They all involve finding $\lim_{n\rightarrow\infty}$ of $a_{n}$, when $a_{n}$ isn't given.

1.) $a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3}$ $(n\geq{1})$
Find $\lim_{n\rightarrow\infty}a_{n}$

2.) $a_{1}=2, a_{n+1}=\frac{2a_{n}+3}{a_{n}+2}$ $(n\geq{1})$
Find $\lim_{n\rightarrow\infty}a_{n}$

3.) Determine the limit of $(a_{n})$, where
$a_{1}=2,a_{n+1}=\sqrt{2+2a_{n}}$ $(n\geq{1})$

If someone could show me how to do one of them I'd really appreciate it!!

Thanks,
Jo

In every case, assume that L is the limit.

Then you can notice that $\lim_{n \to \infty} a_n=\lim_{n \to \infty} a_{n+1}=L$

For the second one, as an example :

$L=\frac{2L+3}{L+2}$

---> $L^2=3$

Then, see if $a_n$ is negative or positive in order to chose the negative or the positive solution :)
• May 18th 2008, 10:05 AM
simplysparklers
Thank you so much Moo!!You rock!!! (Rock) :D
• May 18th 2008, 10:08 AM
galactus
#3, you could try this way.

$\lim_{n\rightarrow{\infty}}\sqrt{2+2a_{n}}$

$L=\sqrt{2+2L}$

$L^{2}-2L-2=0$

$L=\sqrt{3}+1, \;\ L=1-\sqrt{3}$

Reject the negative one because the terms in the sequence are positive.

$\lim_{n\rightarrow{\infty}}a_{n}=1+\sqrt{3}$
• May 18th 2008, 10:10 AM
Isomorphism
Quote:

Originally Posted by simplysparklers
Hey guys!
I have 3 similar problems, if someone could please show me how to do one of them that would be great!
They all involve finding $\lim_{n\rightarrow\infty}$ of $a_{n}$, when $a_{n}$ isn't given.

1.) $a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3}$ $(n\geq{1})$
Find $\lim_{n\rightarrow\infty}a_{n}$

2.) $a_{1}=2, a_{n+1}=\frac{2a_{n}+3}{a_{n}+2}$ $(n\geq{1})$
Find $\lim_{n\rightarrow\infty}a_{n}$

3.) Determine the limit of $(a_{n})$, where
$a_{1}=2,a_{n+1}=\sqrt{2+2a_{n}}$ $(n\geq{1})$

If someone could show me how to do one of them I'd really appreciate it!!

Thanks,
Jo

All extremely similar,

For the first one we have already proved in earlier threads that $a_n$ is monotonically increasing. So let the limit be L, then

$\lim_{n \to \infty} a_{n+1}=\lim_{n \to \infty} \frac{3a_{n}+1}{a_{n}+3} \Rightarrow L = \frac{3L+1}{L+3} \Rightarrow L = \pm1$

But the negative solution is not possible since $a_n$ is monotonic increasing and $a_1 = 0$.

So $L = 1$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For the second one, do a similar thing:

First show by induction $2 \leq a_n < \sqrt 3$
Then show its monotonically increasing. Then finally since the limit exists, do the same ol' trick of letting $\lim_{n \to \infty}a_n = \lim_{n \to \infty} = a_{n+1} = L$ again :)
Then finally choose the positive solution because $2 \leq a_n < \sqrt 3$ :)

EDIT: I am sooo slow :(
• May 18th 2008, 10:17 AM
simplysparklers
Isomorphism got there before me Moo!!But I do get it now, it's actually fairly simple! (Itwasntme)
Thank you guys so much for yer help!!Much love and hugs to ye all!!(Bear)