I'm trying to make sense of this question.
" Find the function x(t) where x(1) = 0 and examine the behavior of x(t) as t -->infinity.
S(2x/2-x^2)dx = S dt " (S is the integral sign)
$\displaystyle \int\frac{2x}{2-x^2}\,dx=\int\,dt$?
Make a u sub: $\displaystyle u=2-x^2\implies-\,du=2x\,dx$.
Thus, we have $\displaystyle -\int\frac{\,du}{u}=\int\,dt\implies -\ln\left|u\right|=t+C\implies \ln\left|2-x^2\right|=-t+C$ $\displaystyle \implies \left|2-x^2\right|=Ce^{-t}\implies 2-x^2=Ke^{-t}\implies x(t)=\sqrt{2-Ke^{-t}}$
Applying the initial condition, we have $\displaystyle 0=\sqrt{2-Ke^{-1}}\implies K=2e$.
Therefore, $\displaystyle x(t)=\sqrt{2-2e^{1-t}}$.
Now, what is $\displaystyle \lim_{t\to\infty}\sqrt{2-2e^{1-t}}$?