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.
and here the tutors don't know how to solve the problems!
