Differentiation Question 118
Question: Let the sequence $ <b _{n}> $ of real numbers satisfies the recurrence relation $ {b _{n+1}}=\frac{1}{3}( 2b _{n}+\frac{125}{b^2 _{n}} ),b _{n}\ne 0. $ Then find $ \underset{n\to \infty }{\mathop{\lim }}b _{n}. $
Options:
A) 10
B) 15
C) 5
D) 25
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let $ \underset{n\to \infty }{\mathop{\lim }}b _{n}=b $ now, $ {b _{n+1}}=\frac{1}{3}( 2b _{n}+\frac{125}{b_n^{2}} ) $ or $ \underset{n\to \infty }{\mathop{\lim }}{b _{n+1}}=\frac{1}{3}( 2\underset{n\to \infty }{\mathop{\lim }}b _{n}+\frac{125}{\underset{n\to \infty }{\mathop{\lim }}b_n^{2}} ) $ or $ b=\frac{1}{3}( 2b+\frac{125}{b^{2}} ) $
$ \Rightarrow \frac{b}{3}=\frac{125}{3b^{2}}\Rightarrow b^{3}=125orb=5. $
BETA
