# Thread: Three similar limit problems

1. ## 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 $\displaystyle \lim_{n\rightarrow\infty}$ of $\displaystyle a_{n}$, when $\displaystyle a_{n}$ isn't given.

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

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

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

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

Thanks,
Jo

2. Hello,

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 $\displaystyle \lim_{n\rightarrow\infty}$ of $\displaystyle a_{n}$, when $\displaystyle a_{n}$ isn't given.

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

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

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

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 $\displaystyle \lim_{n \to \infty} a_n=\lim_{n \to \infty} a_{n+1}=L$

For the second one, as an example :

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

---> $\displaystyle L^2=3$

Then, see if $\displaystyle a_n$ is negative or positive in order to chose the negative or the positive solution

3. Thank you so much Moo!!You rock!!!

4. #3, you could try this way.

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

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

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

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

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

$\displaystyle \lim_{n\rightarrow{\infty}}a_{n}=1+\sqrt{3}$

5. 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 $\displaystyle \lim_{n\rightarrow\infty}$ of $\displaystyle a_{n}$, when $\displaystyle a_{n}$ isn't given.

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

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

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

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 $\displaystyle a_n$ is monotonically increasing. So let the limit be L, then

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

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

So $\displaystyle L = 1$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For the second one, do a similar thing:

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

EDIT: I am sooo slow

6. Isomorphism got there before me Moo!!But I do get it now, it's actually fairly simple!
Thank you guys so much for yer help!!Much love and hugs to ye all!!