Prove that $\forall \epsilon > 0$: $\lim_{t\to\infty}t^{-2}\int_{0}^{t}[(f(x))^{1+\epsilon}/f'(x)]\,\mathrm dx =+\infty$ 
Let $f: [0,+\infty) \to [0,+\infty)$ be differentiable, $f' > 0$. Prove that $$\forall \epsilon > 0: \lim_{t\to\infty}\dfrac{1}{t^2}\int_{0}^{t}\dfrac{\left(f(x)\right)^{1+\epsilon}}{f'(x)}\mathrm dx =+\infty$$

I used L'Hôpital's rule and got 
$$\lim\limits_{t\to\infty}\dfrac{1}{t^2}\int_{0}^{t}\dfrac{\left(f(x)\right)^{1+\epsilon}}{f'(x)}\mathrm dx =\dfrac{1}{2}\displaystyle\lim_{t\to\infty}\dfrac{\left(f(t)\right)^{1+\epsilon}}{tf'(t)}$$
If $\lim\limits_{t\to\infty}f'(t)$ exists, then we can prove the above statement using L'Hôpital's rule. But if $\lim\limits_{t\to\infty}f'(t)$ does not exist, I don't know how to proceed
 A: 
Partial solution. It is easy to see that $ f $ has an inverse $ g $. We will solve the problem with the additional hypothesis that $ g $ is a polynomial function.

For the inverse function theorem, since $f^\prime(x)> 0$ for all $ x \in [0, \infty) $ then $f:[0,\infty)\to [f(0),\infty)$ has an inverse global $g:[f(0),\infty)\to [0,\infty)$. Notice that we consider $ f^\prime(0) $ as is derivable at right of zero. In addition we have the equalities
$$ 
\begin{matrix}
g(f(x))=x \quad & \quad f(g(y))=y \quad & \quad g(y)=x \quad & \quad f(x)=y\\
\\
g^\prime(y)=\frac{1}{f^\prime (x)} \quad & \quad g^\prime(y)=\frac{1}{f^\prime (g(y))}
\quad & \quad g(s)=t \quad & \quad f(t)=s
\end{matrix}
$$
By the formula of change of variables we have
\begin{align}
\int_0^{t} \big( f(x) \big)^{1+\epsilon}\frac{1}{f^\prime(x)} \mathrm{d}x
=&
\int_{g(f(0))}^{g(s)} \big( f(x) \big)^{1+\epsilon}\frac{1}{f^\prime(x)} \mathrm{d}x,
\\
=&
\int_{f(0)}^{s} \big( f(g(y)) \big)^{1+\epsilon}\frac{1}{f^\prime(g(y))}g^\prime(y) \mathrm{d}y,
\\
=&
\int_{f(0)}^{s} \big( y \big)^{1+\epsilon}\cdot g^{\prime}(y)\cdot g^{\prime}(y)\mathrm{d}y,
\end{align}
Now, check for every polynomial function $ g(y)=a_ny^n+\ldots+a_1 y+a_0$  that
$$
\Phi_g(s)=
\frac{1}{g(s)^2}\int_{f(0)}^{s} \big( y \big)^{1+\epsilon}\cdot g^{\prime}(y)\cdot g^{\prime}(y)\mathrm{d}y,
=
\frac{1}{t^2}\int_0^{t} \big( f(x) \big)^{1+\epsilon}\frac{1}{f^\prime(x)} \mathrm{d}x
$$
is an function  such that $\lim_{s\to\infty}\Phi_{g}(s)=\infty$.  
