[College Calc 01-02. Limits of Sequences] #04

If it is not given that $\displaystyle \lim_{n \to \infty} a_n$ exists, we can find it as follows:

$\begin{aligned} a_{n+1} & = \frac 12 a_n - 4 \\ a_{n+1} + 8 & = \frac 12(a_n + 8) \end{aligned}$

Let $b_n = a_n + 8 \implies b_{n+1} = \dfrac 12 b_n$ Then the characteristic equation of the linear recurrent relations is $r = \dfrac 12$ and $b_n = c \left(\dfrac 12\right)^n$ , where $c$ is a constant. Since $b_1 = a_1 + 8 = 11 = \dfrac 12 c \implies c = 22$ , $b_n = a_n + 8 = 22\left(\dfrac 12 \right)^n$ , and

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} 22\left(\dfrac 12 \right)^n - 8 = \boxed{-8}$

It is already given in the problem that the sequence converges to a limit -- call it $\alpha.$ Then, we must also have that $\displaystyle\lim_{n\to\infty} a_{n+1} = \alpha.$

From the recurrence relation,

$\alpha = \frac{1}{2}\alpha - 4\implies \alpha = -8.$

Therefore, $\displaystyle\lim_{n\to\infty}a_n=\boxed{-8}.$