A difference equation

The linear recurrence relation $x_{n+2} = 2x_{n+1} + 3x_n$ has the following characteristic equation.

$\begin{aligned} r^2 & = 2r+3 \\ r^2 - 2r - 3 & = 0 \\ (r-3)(r+1) & = 0 \\ \implies x_n & = c_1(3^n) + c_2(-1)^n & \small \color{#3D99F6} \text{Putting }n=0 \\ x_0 & = c_1 + c_2 = 0 & \small \color{#3D99F6} \implies c_2 = -c_1 \\ x_1 & = 3c_1 - c_2 = 4c_1 = 1 & \small \color{#3D99F6} \implies c_1 = \frac 14 \\ \implies x_n & = \frac 14 \left(3^n + (-1)^{n+1}\right) \end{aligned}$

Therefore,

$\begin{aligned} L & = \lim_{n \to \infty} \frac {x_{n+1}}{x_n} \\ & = \lim_{n \to \infty} \frac {3^{n+1} + (-1)^n}{3^n + (-1)^{n+1}} & \small \color{#3D99F6} \text{Divide up and down by }3^n. \\ & = \lim_{n \to \infty} \frac {3 + \frac {(-1)^n}{3^n}}{1 + \frac {(-1)^{n+1}}{3^n}} \\ & = \boxed 3 \end{aligned}$

The above difference equation has the characteristic equation:

$r^2 - 2r - 3 = (r+1)(r-3) = 0 \Rightarrow r = -1,3$

and the sequence $x_{n} = A(-1)^n + B(3)^n$ . Using the initial conditions $x_{0}=0, x_{1}=1$ , we obtain:

$0 = A + B$ and $1 = -A + 3B$

or $A = -\frac{1}{4}, B = \frac{1}{4}$ so that our sequence is described by $x_n = \frac{1}{4} \cdot [3^n + (-1)^{n+1}]$ . For the above limit $\frac{x_{n+1}}{x_{n}}, n \rightarrow \infty$ we have $\frac{ \frac{1}{4} \cdot [3^{n+1} + (-1)^{n+2}] }{ \frac{1}{4} \cdot [3^n + (-1)^{n+1}] }$ . If we let $u = 3^n, v = (-1)^{n}$ , then we have:

$\frac{3u + v}{u-v} = 3 + \frac{4v}{u-v} = 3 + \frac{4(-1)^{n} }{3^n + (-1)^{n+1}}$

As $n \rightarrow \infty$ , the quanitity $3^n + (-1)^{n+1}$ grows much faster than $4(-1)^n$ (which is bounded to $\pm 4$ ). The limit is ultimately equal to $3 + 0 = \boxed{3}.$