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Let the sequence ltb(n)gt of real number...

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 `lim_(ntoo0) b_(n).`

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Verified by Experts

The correct Answer is:
5
Let `underset(ntooo)limb_(n)=b`
Now `b_(n+1)=1/3(2b_(n)+(125)/(b_(n)^(2)))`
`implies" "underset(ntooo)limb_(n+1)=1/3(2underset(ntooo)limb_(n)+(125)/(underset(ntooo)limb_(n)^(2)))`
`implies" "b=1/3(2b+(125)/(b^(2)))" "( :.underset(ntooo)limb_(n)=underset(ntooo)limb_(n+1)=b)`
`implies" "b/3=(125)/(3b^(2))`
`implies" "b^(3)=125`
`implies" "b=5`
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