What must a function satisfy in order to say that a certain limit exists? Suppose that $f:R_+\to R_+$ is $C^2$, and that $f(0)=f'(0)=f''(0)=0$, and that for all $x$, $\frac{xf''(x)}{f'(x)}\geq1$. From here we can safely say that $\lim\inf_{x\to0}\frac{xf''(x)}{f'(x)}\geq1$. But, under what conditions on $f$ can we say that the $\lim_{x\to0}\frac{xf''(x)}{f'(x)}$ exists and is greater than 1? 

First, an observation: the condition $\frac{xf''(x)}{f'(x)}\geq \alpha$ is equivalent to the function $x^{-\alpha} f'(x)$ being nondecreasing, since
$$(x^{-\alpha} f'(x))' =  (x f''(x)-\alpha  f'(x)) x^{-\alpha-1} $$
So, if you want the limit to be greater than $1$, the derivative $f'$ must satisfy $f'(x)=o(x^{1+\epsilon})$ for some $\epsilon>0$. 
However, any bound on $f'$ is insufficient to get the existence of a limit involving $f''$. Hence, the following appears to be the most practical sufficient condition: 
Claim. Let $p>0$. If $\lim_{x\to 0 } x^{-p} f''(x)$ exists and is nonzero, then $\lim_{x\to 0} \frac{xf''(x)}{f'(x)} = 1+p$.
Proof. Let $c = \lim_{x\to 0 } x^{-p} f''(x)$. Integration yields $f'(x) = \frac{c+o(1)}{1+p} x^{1+p}$, hence $$\frac{xf''(x)}{f'(x)} =  \frac{x (c+o(1))x^p}{\frac{c+o(1)}{1+p} x^{1+p}} = 1+p+o(1)$$

And here is a necessary condition: 
Claim. Let $p>0$. If $\lim_{x\to 0} \frac{xf''(x)}{f'(x)} = \alpha$, then $f'(q) = o(x^{\alpha-\epsilon})$ and $x^{\alpha+\epsilon} = o(f'(x))$ for every $\epsilon>0$.
Proof. Let's consider $x>0$ for simplicity. By L'Hospital, 
$$
\lim_{x\to 0} \frac{\log f'(x)}{\log x} = \lim_{x\to 0} \frac{f''(x)/f'(x)}{1/x} = \alpha
$$
Hence $\log f'(x) - (\alpha \pm \epsilon)\log x \to \pm \infty$ as claimed.
