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
$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0$.
 A: 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).
A: 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.
A: Added: Here's a more concise way of phrasing the argument in (II.) and (III.), that is, of completing the proof once it is established that $f$ must be eventually monotonic (see Paramanand's argument, which I replicate in I). If $f$ is eventually monotonic and satisfies $(1)$ but not $(2)$, then $f$ must eventually be bounded away from $0$, and $f'$ must eventually have constant sign. In that case, $(1)$ implies that $\lim_{x\to\infty} f'(x)/|f(x)|^{b/a}$ exists and is equal to either $1$ or $-1$. For $b/a>1$, this can't be true; otherwise, for sufficiently large $A$,
\begin{align*}
\limsup_{x\to\infty} \left|{1\over|f(x)|^{b/a-1}} - {1\over|f(A)|^{b/a-1}}\right| &=  \limsup_{x\to\infty} \left|(-b/a+1)\int_A^x {f'(t)\over |f(t)|^{b/a}}\,dt\right|\\
& = |-b/a+1|\int_A^\infty {|f'(t)|\over |f(t)|^{b/a}}\,dt \\
& \geq |-b/a + 1| \int_A^\infty {1\over2}\,dt,
\end{align*}
say. The last quantity is infinite, but the first must be finite since $f$ is bounded away from $0$, and this is our contradiction. (Like I said, this is basically a rephrasing of the argument below.)

Here is I think the (or at least a) proper generalization, together with a complete proof (borrowing, in places, from Paramanand's answer): If $f$ is differentiable on $[0,\infty)$ and there exist $a$ and $b$ with $0<a<b$ such that
\begin{align*}
 \lim_{x\to\infty}{|f'(x)|^a - |f(x)|^b} = 0, \tag{1}
\end{align*}
then
\begin{align*}
 \lim_{x\to\infty}f'(x) = \lim_{x\to\infty} f(x) = 0. \tag{2}
\end{align*}
This resolves China Math's problem, since $\left|f'(x)^2 + f(x)^3\right|\geq \left||f'(x)|^2-|f(x)|^3\right|$, and so we can take $a = 2$ and $b = 3$. I suspect that the statement is sharp in the sense that, whenever $0< b\leq a$, there exist functions that satisfy $(1)$ but not $(2)$ (mike's answer suggests that this is true for $b<a$ and the exponential function shows that this is certainly true when $a = b$, though it is also true that $(2)$ is satisfied when $a = b = 1$ if the quantity in $(1)$ is replaced with $f'(x) + f(x)$), but I haven't really attempted to construct them. I suppose you could also ask what happens when $a$ and $b$ are permitted to be negative, but I'm tired.
Anyway, to prove this, I'll follow Paramanand's lead to show (I.) that a function satisfying $(1)$ but not $(2)$ must be eventually monotonic; (II.) that such a function must be unbounded; and (III.) that no function satisfying $(1)$ can be unbounded. This last step—that no unbounded, monotonic function can satisfy $(1)$ but not $(2)$—is the only thing that Paramanand didn't get around to proving, and it basically boils down to the proposition that there is no unbounded, monotonic function $f$ eventually satisfying $|f'|>|f|^\nu$ for some $\nu>1$. Apologies for the recitation of Paramanand's arguments toward the beginning, but I wanted to be complete.
I. If $f$ is not eventually monotonic, then, as Paramanand has pointed out, the set of points, call it $E$, at which it attains a local extremum is unbounded. Since $f'$ vanishes at each point of $E$, we may conclude that $f(x)\to0$ as $x\to\infty$ through $E$. But $\limsup_{x\to\infty}|f(x)| = \limsup_{x\in E, x\to\infty} |f(x)|$, and so it must be that, if $f$ is not eventually monotonic, $f(x)\to0$ as $x\to\infty$.
II. We may now suppose $f$ to be monotonic. If it is bounded, then $f(x)$ does tend to a limit, call it $L$, as $x\to\infty$. The equation $(1)$ then implies that $|f'(x)|$ converges to $|L|$ as $x\to\infty$. A variation on Paramanand's mean value argument then establishes that $L = 0$:
\begin{align*}
 \text{$|f(x+1) - f(x)| = |f'(\xi)|$ for some $\xi\in(x,x+1)$};
\end{align*}
as $x\to\infty$, the left side of the equation vanishes while the right side necessarily tends to $L$. Thus $L = 0$, and we are done in case $f$ is bounded.
III. Finally, assume that $f$ is unbounded. (We also retain, of course, the assumption that $f$ is monotonic.) Then $|f(x)|\to\infty$ and $f$ does not vanish for sufficiently large $x$—I'm going to show that this cannot happen when the growth of $f'$ is a lot greater than that of $f$. Since $0<a<b$, we may choose $\nu$ with $1<\nu < b/a$. I claim now that $|f'(x)|>|f(x)|^\nu$ for all sufficiently large $x$. Otherwise, the set of $x$ for which
\begin{align*}
 |f'(x)|^a - |f(x)|^b < |f(x)|^{a\nu} - |f(x)|^b
\end{align*}
would be unbounded, and that is not so because the left hand side is supposed to vanish in the limit $x\to\infty$, while the right hand side tends to $-\infty$ by virtue of the facts that $|f(x)|\to\infty$ with $x$ and $0 < a\nu < b$. My claim is justified.
But now consider the function $g(x) = |f(x)|^{-\nu+1}$. Because $f(x)$ doesn't vanish for sufficiently large $x$, $g$ is differentiable for sufficiently large $x$, and because $\nu>1$ and $|f(x)|\to\infty$ with $x$, we can be sure that $g(x)\to0$ as $x\to\infty$. We can also be sure, however, that for $x$ sufficiently large (large enough that $g$ is differentiable at $x$ and $|f'(x)|>|f(x)|^\nu$)
\begin{align*}
 |g'(x)| = |(-\nu+1)f(x)^{-\nu}f'(x)| > |-\nu + 1|.
\end{align*}
This, together with the mean value theorem, raises a problem: if $x$ and $y$ are sufficiently large, as above, then, for some $\xi\in (x,y)$,
\begin{align*}
 |g(x) - g(y)| & = |g'(\xi)(x-y)| \\
 & > |-\nu + 1||x-y|.
\end{align*}
Since $|-\nu+1|>0$, the right hand side tends to infinity with $x$ if we hold $y$ fixed. That is the desired contradiction, because we should have $\lim_{x\to\infty}g(x) = 0$ and hence the left hand side should tend in the limit $x\to\infty$ to the finite quantity $|g(y)|$.
