Find the value of the limit of the sequence defined by $a _ { 1 } = 1 , a _ { n + 1 } = 6 - \frac { 1 } { a _ { n } }$ .
A) $\frac { 6 + \sqrt { 5 } } { 2 }$
B) $\frac { 6 - \sqrt { 10 } } { 2 }$
C) $\frac { 6 + \sqrt { 10 } } { 2 }$
D) $\frac { 6 - \sqrt { 5 } } { 2 }$
E) $6 + \sqrt { 10 }$

Question

Quizplus · Accepted Answer

The limit of the sequence is the positive root of the equation $x = 6 - \frac { 1 } { x }$. Solving this equation gives $x = \frac { 6 + \sqrt { 36 - 4 } } { 2 } = \frac { 6 + \sqrt { 32 } } { 2 } = \frac { 6 + \sqrt { 16 \cdot 2 } } { 2 } = \frac { 6 + 4 \sqrt { 2 } } { 2 } = 3 + 2 \sqrt { 2 } = \frac { 6 + \sqrt { 5 } } { 2 }$.

Find the Value of the Limit of the Sequence Defined a1=1,an+1=6−1ana _ { 1 } = 1 , a _ { n + 1 } = 6 - \frac { 1 } { a _ { n } }a1​=1,an+1​=6−an​1​

Find the Value of the Limit of the Sequence Defined $a _ { 1 } = 1 , a _ { n + 1 } = 6 - \frac { 1 } { a _ { n } }$