Let the sequence$< bn >$of real numbers satisfy; the recurrence relation $b_{n+1}=\frac{1}{3}\left(2b_n+\frac{125}{b_{n^2}}\right),b_n\ne 0,\text{ then }\underset{n\to \mathrm{\infty }}{lim} b_n=$

Let $\underset{n\to \mathrm{\infty }}{lim} b_n=b$ then

$\begin{array}{l}{b}_{n+1}=\frac{1}{3}\left(2b_n+\frac{125}{b_n^2}\right)\\ \Rightarrow \underset{n\to \mathrm{\infty }}{lim} b_{n+1}=\frac{1}{3}\left\{2\underset{n\to \mathrm{\infty }}{lim} b_n+\frac{125}{\underset{n\to \mathrm{\infty }}{lim} b_n^2}\right\}\\ \Rightarrow b=\frac{1}{3}\left\{2b+\frac{125}{b^2}\right\}\\ \Rightarrow \frac{b}{3}=\frac{125}{3b^2}\Rightarrow b^3=125\Rightarrow b=5\end{array}$