# Character of a solution to a Cauchy problem

Consider $f:\mathbb{R}\rightarrow \mathbb{R}$ in $C^1$ s.t. $f'\geq 0$ for every $x\in\mathbb{R}$ and $f(0)=0$. I have proved that the solution of the problem $$\begin{cases} \dot{x}=\frac{1}{1+tf(x)} \\ x(0)=0 \end{cases}$$ exists on the whole $\mathbb{R}$ but I can't show that $$\lim_{x\rightarrow +\infty} x(t) = +\infty.$$

You basically want to show that for all $M>0$, the inequality $x(t)>M$ is eventually true. Note that $f(s)>0$ for all $s>0$, so $x(t)$ is monotonic ascending in $t$. Thus, it's enough to show that the inequality $x(t)\le M$ does not hold for all times.
Now, take some $M>0$ and suppose (heading toward contradiction) that $x(t) \le M$ for all times $t>0$. The function $f$ monotonically non-decreases (as its derivative is non-negative), so for all times $t$,
$$\dot{x}(t)=\frac{1}{1+tf(x)}\ge\frac{1}{1+f(M)t}$$
as $f(x)\le f(M)$ by our assumption. Note that $f(M)$ is just a constant, and that $x(0)=0$ so we get that:
$$M\ge \limsup_{t\rightarrow\infty}(x(t)-x(0))=\limsup_{t\rightarrow\infty}\int_0^t \dot{x}(s)ds\ge$$ $$\ge \limsup_{t\rightarrow\infty}\int_0^t \frac{ds}{1+f(M)s}=\limsup_{t\rightarrow\infty}\log(1+f(M)t)=\infty$$