The solution to the Landau equation
$$\frac{d \vert A\rvert^2}{d t}=2\sigma \lvert A \rvert^2-l \lvert A \rvert^4$$
is
$$\lvert A \rvert ^2= \frac{A_0^2}{{\frac{l}{2\sigma} A_0^2+\left(1-\frac{l}{2 \sigma} A_0^2 \right) e^{-2\sigma t} }},$$ where $l$ and $\sigma$ are constant, $A_0$ is the initial value of $\lvert A \rvert$.
If $l>0$ and $\sigma>0$, the textbook showed that the above solution gives
$$\lvert A \rvert \sim A_0 e^{\sigma t}$$ as $t\to-\infty$ and $A_0\to 0$. But I think simply $$\lim _{t \to -\infty , A_0 \to 0} \lvert A \rvert ^2=\frac{0}{\infty}=0$$
Can anybody suggest me how to show this asymptotic relation? Thank you in advance!
Note \begin{eqnarray} \lvert A \rvert ^2 &=& \frac{A_0^2}{{\frac{l}{2\sigma} A_0^2+\left(1-\frac{l}{2 \sigma} A_0^2 \right) e^{-2\sigma t} }}\\ &=& \frac{A_0^2e^{2\sigma t}}{{\frac{l}{2\sigma} A_0^2e^{2\sigma t}+\left(1-\frac{l}{2 \sigma} A_0^2 \right) }} \end{eqnarray} and hence \begin{eqnarray} \lim_{t\to-\infty}\frac{\lvert A \rvert ^2}{e^{2\sigma t}} &=& \lim_{t\to-\infty}\frac{A_0^2}{{\frac{l}{2\sigma} A_0^2e^{2\sigma t}+\left(1-\frac{l}{2 \sigma} A_0^2 \right) }}=A_0^2. \end{eqnarray} So $|A|\sim A_0e^{\sigma t}$ as $t\to-\infty$.