oK, I'll try to help you on the 3rd one. In order to be able to find [tex]\lim_{n\rightarrow \infty}x_n[/tex] you first need to show that the sequence
[tex]x_{n+1}=\frac{3+2x_{n}}{3+x_n}[/tex] 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
[tex]x_1=\frac{3+2x_o}{3+x_o}=\frac{3+2}{3+1}=\frac{5}{4}[/tex] 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
[tex]x_n>x_{n-1}=>x_n-x_{n-1}>0--------(IH)[/tex] thus we suppose that x_n is increasing.
Now we want to prove that also
[tex]x_{n+1}-x_n>0 (?)[/tex]
that is:
[tex]\frac{3+2x_{n+1}}{3+x_n}-\frac{3+2x_{n-1}}{3+x_{n-1}}=...>0[/tex] You do the calculation.
SO, this means that [tex]x_n[/tex] is a monotono increasing sequence.
Now our job is to prove that it is upper bounded.
Ok, here it is what we do
[tex]x_{n+1}=\frac{3+2x_{n}}{3+x_n}<\frac{3+2x_n}{x_n}=\frac{3}{x_n}+2<5[/tex] 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
[tex]\lim_{n\rightarrow \infty}x_n=L[/tex] Now,
[tex]\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}[/tex]
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!