# Limit Sequence !

1. Oct 10, 2008

### Nanie

Hi! I want to know if I do correctly this excercise!!!!!!!!!!!!!????:grumpy:

I am in stress because of this class!!!

Me está volviendo loca!!!!!!!:yuck:

2. Oct 10, 2008

### Nanie

The first one, the proffesor gave us a hint and I do that....but these two I have no idea how to start....

I will really apreciate if anyone can tell me something, maybe just how to started it or what should I use, I tried to find some help in the university but they dont know how to do it.
And sorry if I post many topics in the forum but really I need som help

The proffesor gave us this kind of challenge of excercise :rofl: and here the tutors don't know how to solve the problems!:rofl::rofl:

3. Oct 10, 2008

### sutupidmath

oK, I'll try to help you on the 3rd one. In order to be able to find $$\lim_{n\rightarrow \infty}x_n$$ you first need to show that the sequence

$$x_{n+1}=\frac{3+2x_{n}}{3+x_n}$$ is monotonic and bounded.

To show that it is monotonic( monotono increasing/decreasing) try to play around a little bit, by finding values, for n=0, n=1, n=2,

for example, we know that Xo=1.
now let n=0, and we get

$$x_1=\frac{3+2x_o}{3+x_o}=\frac{3+2}{3+1}=\frac{5}{4}$$ and also try for n=1, etc.

What do you see, it looks like that the sequence is increasing doesn't it?

Now, this is sufficient to try to use induction to prove it in general. THus we suppose that

$$x_n>x_{n-1}=>x_n-x_{n-1}>0--------(IH)$$ thus we suppose that x_n is increasing.
Now we want to prove that also

$$x_{n+1}-x_n>0 (?)$$
that is:

$$\frac{3+2x_{n+1}}{3+x_n}-\frac{3+2x_{n-1}}{3+x_{n-1}}=.........>0$$ You do the calculation.

SO, this means that $$x_n$$ is a monotono increasing sequence.

Now our job is to prove that it is upper bounded.

Ok, here it is what we do

$$x_{n+1}=\frac{3+2x_{n}}{3+x_n}<\frac{3+2x_n}{x_n}=\frac{3}{x_n}+2<5$$ since n>0, Xo=1, and since x_n is increasing.

Now, since the sequence is bounded and monotonic, we know that it converges somewhere. so let

$$\lim_{n\rightarrow \infty}x_n=L$$ Now,

$$\lim_{n\rightarrow \infty} x_{n+1}=\frac{3+2 \lim_{n\rightarrow \infty}x_{n}}{3+\lim_{n\rightarrow \infty}x_n}=>L=\frac{3+2L}{3+L}$$

Now, all you need to do is solve for L, and interpret your answer.

Last edited: Oct 10, 2008