Let `f:R to R` be a differentiable function with `f(0)=0`. If `y=f(x)` satisfies the differential equation
`(dy)/(dx)=(2+5y)(5y-2)`, then the value of `lim_(x to oo) f(x)` is……………….

Explanation

To solve the given differential equation and find the limit of \( f(x) \) as \( x \) approaches infinity, we will follow these steps:

### Step 1: Rewrite the Differential Equation
The differential equation given is:

\[ \frac{dy}{dx} = (2 + 5y)(5y - 2) \]

### Step 2: Separate Variables
We can separate the variables \( y \) and \( x \):

\[ \frac{dy}{(2 + 5y)(5y - 2)} = dx \]

### Step 3: Integrate Both Sides
Next, we will integrate both sides. To do this, we need to perform partial fraction decomposition on the left-hand side:

\[ \frac{1}{(2 + 5y)(5y - 2)} = \frac{A}{2 + 5y} + \frac{B}{5y - 2} \]

Multiplying through by the denominator:

\[ 1 = A(5y - 2) + B(2 + 5y) \]

Expanding and rearranging gives:

\[ 1 = (5A + 5B)y + (-2A + 2B) \]

Setting up the system of equations:

1. \( 5A + 5B = 0 \)
2. \( -2A + 2B = 1 \)

From the first equation, we can express \( B \) in terms of \( A \):

\[ B = -A \]

Substituting into the second equation:

\[ -2A + 2(-A) = 1 \implies -4A = 1 \implies A = -\frac{1}{4}, \quad B = \frac{1}{4} \]

Thus, we rewrite the integral:

\[ \int \left( -\frac{1}{4(2 + 5y)} + \frac{1}{4(5y - 2)} \right) dy = \int dx \]

### Step 4: Integrate
Now we integrate:

\[ -\frac{1}{4} \int \frac{1}{2 + 5y} dy + \frac{1}{4} \int \frac{1}{5y - 2} dy = x + C \]

The integrals yield:

\[ -\frac{1}{20} \ln |2 + 5y| + \frac{1}{20} \ln |5y - 2| = x + C \]

Combining the logarithms:

\[ \frac{1}{20} \ln \left| \frac{5y - 2}{2 + 5y} \right| = x + C \]

### Step 5: Exponentiate
Exponentiating both sides gives:

\[ \left| \frac{5y - 2}{2 + 5y} \right| = e^{20(x + C)} = e^{20x} e^{20C} \]

Let \( K = e^{20C} \):

\[ \frac{5y - 2}{2 + 5y} = K e^{20x} \]

### Step 6: Solve for \( y \)
Solving for \( y \):

\[ 5y - 2 = K e^{20x} (2 + 5y) \]

Rearranging gives:

\[ 5y - K e^{20x} \cdot 5y = K e^{20x} \cdot 2 + 2 \]

Factoring out \( y \):

\[ y(5 - 5K e^{20x}) = K e^{20x} \cdot 2 + 2 \]

Thus,

\[ y = \frac{K e^{20x} \cdot 2 + 2}{5(1 - K e^{20x})} \]

### Step 7: Find the Limit as \( x \to \infty \)
As \( x \to \infty \), \( K e^{20x} \to \infty \) (assuming \( K > 0 \)), thus:

\[ y \to \frac{K e^{20x} \cdot 2}{-5K e^{20x}} = -\frac{2}{5} \]

However, since we know \( f(0) = 0 \), we need to consider the behavior of the function. The function \( f(x) \) must approach a constant value as \( x \to \infty \).

### Conclusion
The limit of \( f(x) \) as \( x \to \infty \) is:

\[ \lim_{x \to \infty} f(x) = \frac{2}{5} \]