Let the sequence `ltb_(n)gt` of real numbers satisfy the recurrence relation `b_(n+1)=1/3(2b_(n)+(125)/(b_(n)^(2))),b_(n)ne0.` Then find `underset(ntoo0)limb_(n).`

Options

A. 0
B. `oo`
C. 5
D. `2//3`

Answer

C

Explanation

Let `lim_(nto oo)b_(n)=b`. Then,
` b_(n+1)=(1)/(3)(2b_n+(125))/(b_(n^(2))), b_(n) ne 0`, then, `lim_(nto oo) b_n=` then, `lim_(nto oo) b_n=`
`rArr lim_(nto oo) b_(n+1)=(1)/(3) {2 lim_(nto oo) b_(n)+(125)/(lim_(nto oo) b_n^(2))}`
` rArr b=(1)/(3) {2b +(125)/(b^2)}`
` rArr (b)/(3)=(125)/(3b^2)rArr b^3=125rArr b=5`