Given the differential equation $$\frac{dy}{dt}=y(y-1)(y-3)$$ I found $y=1$ is a stable point while $y=0$ and $y=3$ are unstable. However, I have to evaluale $$\lim_{t\to\infty}y(t)$$ without solving the equation and given the initial condition $y(0)=2$.
Since $y=1$ is stable and $y=2$ is "near" of $y=1$, this let me conclude $$\lim_{t\to\infty}y(t)=1?$$
And if there were another stable point, for example, $y=4$, the answer of the limit would still be $1$ because $y=2$ is close to it instead of $y=4$?
Take a look at the sign of $\frac{dy}{dt}$. Observe that $$ \frac{dy}{dt} = y(y-1)(y-3) < 0 \quad \forall \, y \in (1,3).$$ It therefore converges towards $y = 1$ for $y(0) \in (1,3)$.