consider the solution $v$ of the following initial value problem \begin{equation} \begin{cases} (r^{N-1}v'(r))'=r^{N-1}f(v(r)),\quad r\in(0,R_M)\\ v(0)=\alpha,\quad v'(0)=0 \end{cases} \end{equation} where $\alpha>0$ and $(0,R_M)$ is the maximal interval of existence of $v$. Then author intend to prove $R_M=+\infty$. And I'm failed to follow him from here.
The author says:
Let assume, to the contrary, that $R_M$ is finite, then: $$\forall r>0,\quad v'r=r^{1-N}\int_0^rs^{N-1}f(v(s))ds$$ and according $f$ satisified \begin{equation} \begin{cases} f(0)=0\\ f(t)>0,\quad t>0 \end{cases} \end{equation} immediately implies $v′(r) > 0$ for $r \in(0,R_M)$ and so that $\lim\limits_{r\to R_M} v(r) = +\infty$ (otherwise we would have also $\lim\limits_{r\to R_M} v'(r)\in(0,R_M)$ and $v$ could be extended beyond $R_M$, as a solution of ODE, contradicting thus the maximality of $R_M$).
I believe that there is a error that we should obtain $\lim\limits_{r\to R_M} v'(r) = +\infty$ instead of the limit of $v(r)$, but I can't understand how we get the contradiction, thanks for comment and anwsers.