Sequential Lemma!

Relevant wiki: L'Hopital's Rule - Basic

Let $a_n = b_{n-1}$ , then we have $a_1 = b_0 = 4$ , $a_2 = b_1 = 3$ and $2b_{n+1} = 3b_n - b_{n-1}$ for all $n \ge 1$ . Then the characteristic of the linear recurrence relation is as follows:

$\begin{aligned} 2r^2 - 3r + 1 & = 0 \\ (2r-1)(r-1) & = 0 \end{aligned}$

$\begin{aligned} \implies b_n & = \frac {c_1}{2^n} + c_2 \\ b_0 & = c_1 + c_2 = 4 \\ b_1 & = \frac {c_1}2 + C_2 = 3 \\ \implies c_1 & = c_2 = 2 \\ b_n & = 2 + \frac 1{2^{n-1}} \\ \implies a_n & = 2 + \frac 1{2^{n-2}} & \small \color{#3D99F6} \text{Note that }\lim_{n \to \infty} a_n = 2 \end{aligned}$

Therefore,

$\begin{aligned} L & = \lim_{n \to \infty} \frac {\left(a_n-1\right)\left(a_n-2\right)}{\left(\sqrt{a_n+2}-2\right)\left(\sqrt{12a_n-8}-2\right)} & \small \color{#3D99F6} \text{Let } x = a_n \\ & = \lim_{x \to 2} \frac {(x-1)(x-2)}{\left(\sqrt{x+2}-2\right)\left(\sqrt{12x-8}-2\right)} & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 2} \frac {(x-2)+(x-1)}{\frac {\sqrt{12x-8}-2}{2\sqrt{x+2}}+\frac {6\left(\sqrt{x+2}-2\right)}{\sqrt{12x-8}}} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \frac 1{\frac 24 + 0} = \boxed{2} \end{aligned}$

Claim: The explicity expression for the sequence is $a_n = 2 + 4\cdot (\tfrac12)^n.$ Derivation: In general, the solutions of this type of recurrence relations are linear combinations of $a_n = p^n$ . Substituting this, we see that $p^2 = 3p - 1,$ the solutions to which are $p = 1$ and $p = \tfrac12$ . Therefore the general solution of the recurrence relation is $a_n = A + B\cdot (\tfrac12)^n;$ substituting the known values $a_1 = 4$ and $a_2 = 3$ quickly shows that $A = 2$ and $B = 4$ .

To solve the limit, note that $a_n \to 2$ . This takes care of the first term in the numerator and the last term in the denominator; we are left with $\lim_{n\to\infty} \frac{1\cdot 4\cdot (\tfrac12)^n}{\left(2\sqrt{1 + (\tfrac12)^n} - 2\right)\cdot 2};$ both numerator and denominator tend to zero. Use the substitution $\sqrt{1 + x} \approx 1 + \tfrac12x$ for $x \to 0$ to rewrite the limit: $\lim_{n\to\infty} \frac{1\cdot 4\cdot (\tfrac12)^n}{\left(2(1 + \tfrac12\cdot (\tfrac12)^n) - 2\right)\cdot 2} = \frac{4\cdot (\tfrac12)^n}{2\cdot (\tfrac12)^n} = \boxed{2}.$