$a_{n+1}=f(a_n) \overset{?}{\implies}$ the solution of $f(L)=L$ is the limit When there is a recursively defined sequence given by $a_{n+1}=f(a_n)$, if the equation $f(L)=L$ has a unique solution, does it necessarily follow that the sequence is convergent? When the sequence is convergent, does the limit have to be $L$?
Does your answer change if $a_{n+1}=f(a_n,a_{n-1})$ is the recursively defined sequence and $f(L,L)=L$ has a unique solution?
 A: You can easily prove that if $a_n$ converges to $L$ and $f$ is continuous, then $f(L)=L$. In particular, the argument is nearly trivial:
$$L=\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}a_{n+1}=\lim_{n\rightarrow\infty}f(a_n)=f(\lim_{n\rightarrow\infty}a_n)=f(L)$$
where we first note that shifting indices doesn't change the limit, then appeal to the definition of $a_{n+1}$, then finally use $f$'s continuity to exchange it with the limit.
This means that, if $a_n$ converges and $f(L)=L$ has a unique solution, then $a_n$ converges to $L$. In general, if $a_n$ converges, it needs to converge to a solution of that. However, it may diverge, like in functions like $f(x)=2x$ starting with $a_0\neq 0$ or functions like $f(x)=x^2$ start with $|a_0|>1$. One might note that we can still get trouble even over a compact domain - for instance, the function
$$f(x)=\begin{cases}\frac{1}2&\text{if }x\leq \frac{1}4\\2x&\text{if }\frac{1}4\leq x\leq \frac{1}2\\ 2-2x&\text{if }x\geq \frac{1}2\end{cases}$$
defined on $[0,1]$ has that starting a sequence at $a_0=\frac{2}5$ sends us into a periodic repetition of $\frac{2}5$ and $\frac{4}5$ despite having a unique fixed point $f(\frac{2}3)=\frac{2}3$. In fact, even worse, one can show that almost every starting position leads us to the cycle $0,\,\frac{1}2,\,1,\,0,\,\frac{1}2,\,1,\ldots$
Your second question can be addressed using the same machinery by converting it into a function of a single variable; i.e define the auxiliary sequence of pairs $b_n$ as
$$b_n=(a_n,a_{n+1})$$
then note that the function $g$ taking a pair $(x,y)$ to $(f(x),f(y))$ gives us
$$b_{n+1}=g(b_n)$$
and is continuous whenever $f$ was.
A: No.
A simple example is $f(x) = 2x$, and $a_0 = 1$. 
But we do have:
Fact: If $f$ is continuous and the sequence $a_n$ is convergent, and the limit is $L$, then $f(L) = L$.
And also:
Theorem: If $f(x)$ is a continuous function, $f(L) = L$, $a_0 < L$, $f$ is monotone increasing on $[a_0,L]$, $f(a_0) > a_0$, and $L$ is the unique solution to $f(x) = x$ on the interval $[a_0,L]$, then $\lim_{n \rightarrow \infty} a_n = L$. 
Proof: First: The sequence $a_n$ is increasing. By assumption $a_1 = f(a_0) > a_0$. Now assume $a_n > a_{n-1}$. Since $f$ is increasing, $f(a_n) > f(a_{n-1})$, or $a_{n+1} > a_n$. This proves the claim by induction.
Second: The sequence $a_n$ is bounded above by $L$. We know that $a_0 < L$. If $a_n < L$, since $f$ is increasing, we have $f(a_n) < f(L)$, or $a_{n+1} < L$. This proves the claim by induction.
Now the monotone convergence theorem implies that $\lim_{n \rightarrow \infty} a_n$ exists, call it $M$. We also know $M \in [a_0,L]$. Since $f$ is continuous, $f(M) = M$. All this together implies that $M = L$. 
Example: Take $f(x) = \sqrt{x + 2}$, $a_0 = 1$. $f(x)$ is continuous and increasing everywhere, $f$ has a unique fixed point $L = 2$, and $f(a_0) = \sqrt{3} > a_0$. All the conditions of the theorem apply, so the limit of the sequence is $2$.
