How prove that $\lim\limits_{x\to+\infty}f(x)=\lim\limits_{x\to+\infty}f'(x)=0$ if $\lim\limits_{x\to+\infty}([f'(x)]^2+f^3(x))=0$? Question:

Let $f$ be differentiable on $[0,+\infty)$, such as$$\lim_{x\to+\infty}\left([f'(x)]^2+f^3(x)\right)=0$$show that $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0$$

I think this problem is similar to this link and (the  background)
For my problem I guess that if 
$$\lim_{x\to+\infty}\left([f'(x)]^{2m}+f^{2n+1}(x)\right)=0,m,n\in N^{+},n,m>1$$
then we have
$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0$.

This is not a rigorous answer. I just need the space to easily type the equations.
Let us consider the ODE holds near $x\to\infty$:
$$(f'(x))^2+f^3(x)=0 \tag{1}$$
The solution to (1) is:
$$f(x)=-\frac{4}{(x+c)^2} \implies f'(x)=\frac{8}{(x+c)^3}$$
So 
$$\lim_{x\to\infty}f(x)=0,\lim_{x\to\infty}f'(x)=0 \tag{2}$$
Now let us consider the next ODE holds near infinity:
$$(f'(x))^{2m}+f^{2n+1}(x)=0,m,n\in N^{+},n,m>1\tag{3}$$
The solution to (3) is given by:
$$f(x)= 4^{m/(1 - 2m + 2n)}
       \left(\frac{i^{1/m}(-1 + 2m - 2n)(1 + x)}{m}\right)^{(2m)/(-1 + 2m - 2n)}\tag{4}$$
where the integration constant is set to be $i^{1/m}$.
When we set $m=3,n=2$, we obtain:
$$f(x) = -\frac{(1 + x)^6}{46656}, f'(x)=-\frac{(1 + x)^5}{7776} \tag{5}$$
Clearly $f(x),f'(x)$ of (5) do not satisfy (2).
