$\lim_{n \to \infty} \frac{5a_n+6}{a_n-1}=6$

$\lim_{n\to\infty}\frac{5a_n+6}{a_n-1} = 6$

$\frac{5\lim_{n\to\infty} a_n+6}{\lim_{n\to\infty} a_n-1} = 6$

Let $\displaystyle\lim_{n\to\infty}a_n = L$

Thus,
$\frac{5L+6}{L-1} = 6$

$\Rightarrow \boxed{L=12}$

$\lim _{ n\rightarrow \infty }{ \frac { 5{ a }_{ n }+6 }{ { a }_{ n }-1 } =6 } \\ \lim _{ n\rightarrow \infty }{ \frac { 5{ a }_{ n }-5+11 }{ { a }_{ n }-1 } =6 } \\ \lim _{ n_{ n }\rightarrow \infty }{ \frac { 5({ a }_{ n }-1)+11 }{ { a }_{ n }-1 } =6 } \\ \lim _{ n\rightarrow \infty }{ \frac { 5({ a }_{ n }-1) }{ { a }_{ n }-1 } +\frac { 11 }{ { a }_{ n }-1 } =6 } \\ \lim _{ n\rightarrow \infty }{ 5+\frac { 11 }{ { a }_{ n }-1 } =6 } \\ 5+\lim _{ n\rightarrow \infty }{ \frac { 11 }{ { a }_{ n }-1 } =6 } \\ \lim _{ n\rightarrow \infty }{ \frac { 11 }{ { a }_{ n }-1 } =1 } \\ { a }_{ n }-1=11\\ { a }_{ n }=12$

Hello,peace be upon you,

as for the sequence {an}, as it is approaches to infiniy f {an},

5an + 6 / (an -1) = 6

5an + 6 = 6(an -1)

5an = 6an - 6 - 6

5an - 6an = -12

-an = -12

an = 12,

therefore the value of an = 12 as it is approaches to infinity...