To find the limit of the sequence given by [tex]\( a_1 = 3 \)[/tex] and [tex]\( a_{n+1} = \frac{72}{1 + a_n} \)[/tex], we first need to assume that the sequence converges to some limit [tex]\( L \)[/tex]. If the sequence converges, the terms [tex]\( a_n \)[/tex] will approach [tex]\( L \)[/tex] as [tex]\( n \)[/tex] becomes very large. Therefore, we can replace [tex]\( a_n \)[/tex] and [tex]\( a_{n+1} \)[/tex] with [tex]\( L \)[/tex] in the recursive relation.
Given the relation:
[tex]\[ a_{n+1} = \frac{72}{1 + a_n} \][/tex]
As [tex]\( n \to \infty \)[/tex], we assume [tex]\( a_n \to L \)[/tex] and [tex]\( a_{n+1} \to L \)[/tex]. So, we set up the following equation:
[tex]\[ L = \frac{72}{1 + L} \][/tex]
To solve for [tex]\( L \)[/tex], we multiply both sides of the equation by [tex]\( 1 + L \)[/tex] to clear the fraction:
[tex]\[ L(1 + L) = 72 \][/tex]
This leads to the quadratic equation:
[tex]\[ L^2 + L = 72 \][/tex]
We can rearrange this into standard quadratic form:
[tex]\[ L^2 + L - 72 = 0 \][/tex]
Next, we solve this quadratic equation using the quadratic formula [tex]\( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -72 \)[/tex]:
[tex]\[ L = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-72)}}{2 \cdot 1} \][/tex]
[tex]\[ L = \frac{-1 \pm \sqrt{1 + 288}}{2} \][/tex]
[tex]\[ L = \frac{-1 \pm \sqrt{289}}{2} \][/tex]
[tex]\[ L = \frac{-1 \pm 17}{2} \][/tex]
This gives us two potential solutions:
[tex]\[ L = \frac{-1 + 17}{2} = \frac{16}{2} = 8 \][/tex]
[tex]\[ L = \frac{-1 - 17}{2} = \frac{-18}{2} = -9 \][/tex]
Since [tex]\( L = -9 \)[/tex] is not a reasonable limit for our sequence (as the terms [tex]\( a_n \)[/tex] must be positive due to their definition), we discard [tex]\( L = -9 \)[/tex].
Therefore, the limit of the sequence is:
[tex]\[ L = 8 \][/tex]
Hence, the correct answer is:
[tex]\[ \boxed{8} \][/tex]
