How do I show $\liminf_{x \rightarrow \infty} \frac{f^{'}(x)}{f(x)^{a}} \leq 0$ . suppose that $f:(0,\infty) \rightarrow (0,\infty)$ is a differentiable and positive function. Show that for any $a>1$, it must hold that $\liminf_{x \rightarrow \infty} \frac{f^{'}(x)}{f(x)^{a}} \leq 0$.
Can anyone suggest me direction for this question?
 A: Let $\;g:\left(0,+\infty\right)\to\mathbb{R}$ the function defined as
$g(x)=\frac{f(x)^{1-a}}{1-a}\;$ for all $\;x\in \left(0,+\infty\right).$
Since $a>1$ and $f(x)$ is a positive and differentiable function on $\left(0,+\infty\right)$, it follows that $\;g(x)$ is a negative and differentiable function on $\left(0,+\infty\right)$.
Moreover, $\;g’(x)=\frac{f’(x)}{f(x)^a}\;$ for all $\;x\in\left(0,+\infty\right),\;$ therefore $\;\liminf_\limits{x\to +\infty} g’(x)=\liminf_\limits{x\to +\infty} \frac{f’(x)}{f(x)^a}=l\;.$
We have to prove that $\;l\le0\;.$
If $\;l$ were a positive real number, then there would exist $x^*>0$ such that $\;g’(x)>\frac{l}{2}\;$ for all $\;x>x^*.$
Let $\;x^{**}=x^*-\frac{2g(x^*)}{l}>x^*$ and, by applying Lagrange's Mean Value Theorem on $\left[x^*,x^{**}\right]$, there would exist $c\in\left(x^*,x^{**}\right)$ such that
$g(x^{**})=g(x^*)+g’(c)(x^{**}-x^*)>g(x^*)+\frac{l}{2}\left(-\frac{2g(x^*)}{l}\right)=0,$
hence $\;g(x^{**})>0\;,$
but it is a contraddiction because $g(x)$ is a negative function on $\left(0,+\infty\right)$, so it is impossible that $\;l\;$ is a positive real number.
Analogously, $\;l=+\infty$ also leads to a contraddiction.
Hence $\;l\le0$.
