My attempt:
(i) For r=1, $x_n$ = n+1 $\iff$ $\frac{n+1}{1+1/n}$ = $x_{n-1}$ $\iff$ $\frac{n(n+1)}{n+1}$ = $x_{n-1}$ $\iff$ n = $x_{n-1}$
But I don't know how to use what I have shown so far to show that $x_n$ = n+1
(ii) $$\lim_{n\to\infty} x_n = (r+0) \ell = r\ell $$
I know this is the wrong answer, but I'm not sure what else to try.
We have the recursive relationship $x_n=\left(r+\frac1n\right)x_{n-1}$, with initial condition $x_0=1$. For $r=1$, we have $x_n=\left(1+\frac1n\right)x_{n-1}=\left(\frac{n+1}{n}\right)x_{n-1}$.
Thus, we can generate the solution recursively as
$$x_n=\frac{n+1}{n} \frac{n}{n-1} \frac{n-1}{n-2}\cdots \frac{4}{3}\frac32 \frac21=n+1$$
as expected!
Alternatively, we can use induction. We see that $x_1=2$ and $x_2=3$. So, we assume that for some $k$, $x_k=k+1$. Then,
$x_{k+1}=\frac{k+2}{k+1}\times x_k=k+2$
which completes the proof.
For part $(2)$, we assume that $\lim_{n\to \infty}x_n=L$. Then,
$$x_n=\left(r+\frac1n\right)x_{n-1}\implies L=rL\implies r=1\,\,\text{or}\,\,L=0$$
Since we already showed that $x_n$ has no limit if $r=1$, then we conclude that $L=0$.