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).`

Explanation

To solve the problem, we need to find the limit of the sequence \( b_n \) defined by the recurrence relation:

\[ b_{n+1} = \frac{1}{3} \left( 2b_n + \frac{125}{b_n^2} \right) \]

We will denote the limit as \( L \). Assuming that the limit exists, we can write:

\[ L = \frac{1}{3} \left( 2L + \frac{125}{L^2} \right) \]

Now, let's solve for \( L \) step by step.

### Step 1: Set up the equation
We start by substituting \( L \) into the recurrence relation:

\[ L = \frac{1}{3} \left( 2L + \frac{125}{L^2} \right) \]

### Step 2: Multiply both sides by 3
To eliminate the fraction, we multiply both sides by 3:

\[ 3L = 2L + \frac{125}{L^2} \]

### Step 3: Rearrange the equation
Now, we rearrange the equation to isolate terms involving \( L \):

\[ 3L - 2L = \frac{125}{L^2} \]

This simplifies to:

\[ L = \frac{125}{L^2} \]

### Step 4: Cross-multiply
Next, we cross-multiply to eliminate the fraction:

\[ L^3 = 125 \]

### Step 5: Solve for \( L \)
Now, we take the cube root of both sides:

\[ L = \sqrt[3]{125} = 5 \]

### Conclusion
Thus, the limit of the sequence \( b_n \) as \( n \) approaches infinity is:

\[ \lim_{n \to \infty} b_n = 5 \]