Answer
$2$
Work Step by Step
Since, we are given that $a_{n+1}=\dfrac{a_n+6}{a_n+2}$ and $a_1=-1$
Consider $\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\dfrac{a_n+6}{a_n+2}=l$ and $\lim\limits_{n \to \infty} a_n=l$
Thus, $l=\dfrac{l+6}{l+2}$
or, $l^2+l-3=0$
or, $l=-3 $ or, $2$
Since, we have $a_1=-1$, which is $-1 \lt 0$
As per the Principal of Mathematical Induction $a_n \geq 0$, which means that $ l \geq 0$
Hence, the limit of the sequence is $2$
