I want to show that the IVP $$ x^{'}=x^3, \quad x(0)=2 $$ has one solution inside the interval $(-\infty,b)$ for some $b\in\mathbb{R}.$
Using Picard's Thereom, since $f(x(t))=x^3$ is satisfying the Lipschitz condition in the interval $[-B,B]$ with $$ \Big|f(x)-f(y)\Big|=\Big|x^3-y^3\Big|\leq \Big|x^2+xy+y^2\Big|\Big|x-y\Big|\leq 3B^2\Big|x-y\Big|=L\Big|x-y\Big|$$ then there exists a unique solution in some interval $(a,b)$ for some $a,b\in \mathbb{R}$.
Now, to show that this solution can be extended to the left of $a$, we must have $$ \lim_{t\to a_+} x(t)=l\in\mathbb{R} $$ and also $$ (a,l)\in \mathbb{D} $$ with $\mathbb{D}$ the domain of $f(x(t))$.
How can I prove that this limit exists and it's not $\pm \infty$?
You can also solve the problem without finding the solution explicitly. The constant function $0$ is a solution of the equation. By uniqueness of solution, we have $x(t)>0$ for all $t$ where the solution is defined. This implies that $x'>0$, $x$ is increasing and $0<x(t)<2$ for $t<0$ in the interval of existence of the solution. This in turn implies that $x$ is defined all the way up to $-\infty$ and that $\lim_{t\to-\infty}x(t)$ exists. Once you know the limit exists, it is easy to show that it must be $0$.