Let f: ℝ → ℝ be a differentiable function with f(0) = 0 . If y =...

\dfrac{dy}{dx} = 25{y}^{2} - 4$<br />So$\dfrac{dy}{25{y}^{2} - 4} = dx$<br />Integrating$\dfrac{1}{25} \times \dfrac{1}{2 \times \dfrac{2}{5}}\ln \left|\dfrac{y - \dfrac{2}{5}}{y + \dfrac{2}{5}}\right| = x + c$<br />$⇒ \ln \left|\dfrac{5y - 2}{5y + 2}\right| = 20\left(x + c\right)$<br />Now,$c = 0$as$f\left(0\right) = 0$<br />Hence, \left|\dfrac{5y - 2}{5y + 2}\right| = {e}^{\left(20x\right)}$
$\lim_{x \to - \infty} \left|\dfrac{5f\left(x\right) - 2}{5f\left(x\right) + 2}\right| = \lim_{x \to - \infty} {e}^{\left(20x\right)}$
Now,
$RHS = 0 ⇒ \lim_{x \to - \infty} ( 5f( x ) - 2 ) = 0$
$⇒ lim x \to - \infty$ $f ( x )$$= 2 5$