Limit at Infinity for Solution of First Order Differential Equation

The given differential equation is: $\frac{dy}{dx} = (2 + 5y)(5y - 2) = 25y^2 - 4$

We can rewrite this in separable form: $\frac{dy}{25y^2 - 4} = dx$

Integrating both sides: $\int \frac{dy}{(5y - 2)(5y + 2)} = \int dx$

Using partial fraction decomposition, we have: $\frac{1}{4} \int \left( \frac{1}{5y - 2} - \frac{1}{5y + 2} \right) dy = \int dx$

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

Given the initial condition $f(0) = 0$, i.e., $y = 0$ at $x = 0$: $\frac{1}{20} \ln \left| \frac{-2}{2} \right| = 0 + C \implies C = 0$

Thus, the implicit solution is: $\frac{1}{20} \ln \left| \frac{5y - 2}{5y + 2} \right| = x$

Since $y(0) = 0$, for values near $x = 0$, $-\frac{2}{5} < y < \frac{2}{5}$, which implies $\frac{5y - 2}{5y + 2} < 0$. Taking the absolute value gives: $\ln \left( \frac{2 - 5y}{2 + 5y} \right) = 20x$

Taking the exponential on both sides: $\frac{2 - 5y}{2 + 5y} = e^{20x}$

Solving explicitly for $y = f(x)$: $2 - 5y = (2 + 5y)e^{20x}$ $2(1 - e^{20x}) = 5y(1 + e^{20x})$ $f(x) = \frac{2}{5} \left( \frac{1 - e^{20x}}{1 + e^{20x}} \right)$

Now, evaluating the limit as $x \to -\infty$: $\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{2}{5} \left( \frac{1 - e^{20x}}{1 + e^{20x}} \right)$

Since $\lim_{x \to -\infty} e^{20x} = 0$: $\lim_{x \to -\infty} f(x) = \frac{2}{5} \left( \frac{1 - 0}{1 + 0} \right) = \frac{2}{5} = 0.4$