Limit of solution of differential equation without solving the equation. Given
$$x'(t)=A-B\left(x(t)\right)^2, \quad x(0)=0.$$
Is it possible to find $\lim\limits_{t\to\infty}x(t)$ without solving the differential equation?
Assuming $\lim\limits_{t\to\infty}x'(t)=0$ gives $\lim\limits_{t\to\infty}x(t)=\sqrt{A/B}$ which is correct, but I can not manage to prove that the limit of the derivative is 0.
If it is of any help, the solution to the differential equations is
$$x(t)=\sqrt{\frac AB}\tanh\left(\sqrt{AB}t\right).$$
 A: You can use direction field. When $x=\sqrt{\frac{A}{B}}$, $x'=0$. Then show that it is the only stable equilibrium point so that $x\rightarrow\sqrt{\frac{A}{B}}$ as $t\rightarrow\infty$ with any initial value. 
If $x>\sqrt{\frac{A}{B}}$, $A-Bx^2 <0$. If $-\sqrt{\frac{A}{B}}<x<\sqrt{\frac{A}{B}}$, $A-Bx^2>0$. This shows that $x=\sqrt{\frac{A}{B}}$ is a stable equilibrium. 
A: I was slower than @KittyL, but since I had drawn the actual figure, I'd like to add it. You can see this answer as a supplement to what KittyL writes, and I suggest you to accept KittyL's answer rather than mine.
In the figure/example I have let $A=1$ and $B=1$. What is drawn is the vector field $(1,1-x^2)$. We see that if we start a solution $x(0)>-\sqrt{A/B}=-1$ (this is your case, since $x(0)=0$) then it will tend to the stable equilibrium $\sqrt{A/B}=+1$. If we start with $x(0)<-\sqrt{A/B}$ the solution $x(t)$ will tend to $-\infty$ in finite time.
This is typically part of the subject of a first course in Ordinary differential equations.

