Question

satisfies the recurrence relation $b_{n+1}=\frac{1}{3}(2b_n+\frac{125}{b_n^2}),b_n\ne 0$

Answer 1

$\begin{array}{c}{b}_{n+1}=\frac{1}{3}\left(2b_n+\frac{125}{{b_n}^2}\right)\\ {lim}_{n\to \infty }{b}_n=?\\ So,let{lim}_{n\to \infty }{b}_n={lim}_{n\to \infty }{b}_{n+1}=t\\ Ifn\to \infty b_n\to b_{\infty }\&b_{n+1}\to b_{\infty +1}\cong b_{\infty }\end{array}$

Taking hint both side, we get

$\begin{array}{c}{lim}_{n\to \infty }{b}_{n+1}=\frac{1}{3}\left(2{lim}_{n\to \infty }{b}_n+\frac{125}{{lim}_{n\to \infty }{b_n}^2}\right)\\ t=\frac{1}{3}\left[2t+\frac{125}{t^2}\right]\\ t=\frac{1}{3}\left[\frac{2t^3+125}{t^2}\right]\\ 3t^2=2t^2+125\\ t^3=125\\ t=5\\ So,{lim}_{n\to \infty }{b}_n=t=5\end{array}$