Answer
$2$
Work Step by Step
Given: $a_{n+1}=\dfrac{a_n+6}{a_n+2}$ and $a_1=-1$
Suppose $l=\lim\limits_{n \to \infty} a_n$; $l=\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\dfrac{a_n+6}{a_n+2}$
Thus, $l=\dfrac{l+6}{l+2}$
or, $l^2+l-6=0 \implies (l+3)(l-2)=0$
Then, $l=-3 $ or, $2$
As per the Principle of Mathematical Induction $a_n \geq 0$ for $n\geq 1$, this means that $ l \geq 0$.
Thus, the limit of the sequence is $l=2$.
