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
$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0$.

Let's try in the following manner (a style taken from Hardy's Pure Mathematics for the problem related to $f'(x) + f(x) \to 0$).
Suppose $f'(x) \geq 0$ for sufficiently large $x$. Then $f(x)$ is increasing and hence tends to a limit or to $\infty$. It follows that $f(x)$ has a constant sign for sufficiently large $x$. Since $\{f'(x)\}^{2} + \{f(x)\}^{3} \to 0$, it follows that $f(x)$ must be negative for large $x$. It follows that $f(x) \to A$ where $A \leq 0$. This means that $f'(x) \to (-A)^{3/2}$. When both $f'(x)$ and $f(x)$ tend to a limit then $f'(x) \to 0$ (this is easy to see if consider the equation $f(x) - f(x/2) = (x/2)f'(c)$ by MVT). Thus we see that $A = 0$ and we are done.
Same way we can handle the case when $f'(x) < 0$ for sufficiently large $x$.
If $f'(x)$ changes sign for all sufficiently large $x$ (meaning that given any real number $N > 0$, there is an $x_{1} > N$ such that $f'(x_{1}) < 0$ and an $x_{2} > N$ such that $f'(x_{2}) > 0$) then this means that there are sufficiently large values of $x$ which serve as extrema points of $f$. Since $\left[\{f(x)\}^{3}\right]' = 3\{f(x)\}^{2}f'(x)$ it follows that these points are also the extrema of $\{f(x)\}^{3}$. At these extrema points we have $f'(x) = 0$ and also $\{f'(x)\}^{2} + \{f(x)\}^{3} \to 0$ so that $f(x) \to 0$ when $x$ is tending to $\infty$ through these extrema points. Since the values of $f$ at the extrema points tends to $0$ then the values of $f$ at other large points $x$ will also tend to $0$. It follows that $f(x) \to 0$. And then from condition $\{f'(x)\}^{2} + \{f(x)\}^{3} \to 0$ we get $f'(x) \to 0$.
Update: I think we can't handle the case when $f'(x) < 0$ for sufficiently large $x$. Because in this case it is possible to have $f(x)$ decreasing to $-\infty$ and $f'(x)$ also decreasing to $-\infty$ and thereby lead to a case when $\{f'(x)\}^{2} \to \infty$ and $\{f(x)\}^{3} \to -\infty$ in such a way as to have the sum tending to $0$. I guess we need to tackle this case in some other manner or there exists a counterexample to the question in this case. Hence this should be considered a partial answer.
