Question about asymptotic order How does $\frac{n^{1-2/p}h_{n}^r}{\log(n)}\rightarrow \infty$, $n \rightarrow \infty$ and $h_{n}\rightarrow 0$ ($h_{n}$ is a function of $n$) imply $\frac{(\log(n))^{1/2}}{(nh_{n}^{r+2d})^{1/2}}\rightarrow 0$, where $\log(n)$ is the natural log of $n$ and $r,d,p$ are integers satisfying $r\geq 1, d \geq 2, p \geq 2$?
 A: Since you have specified that $h_n$ is a function of $n$, I'm going to change the notation to refer to this as $h(n)$.  We can write the quantity of interest (without worrying about the square root) as:
$$\frac{\log(n)}{n h(n)^{r+2d}}
= \frac{\log(n)}{n^{1-2/p} h(n)^r} \Bigg/ (n h(n)^{dp})^{2/p}.$$
We can let $L \equiv \lim_{n \rightarrow \infty} n h(n)^{dp}$ denote the limit of the bracketed term (assuming this exists) and note that this limit may be infinite.  If $h$ is continuous then we have:
$$\lim_{n \rightarrow \infty} \frac{\log(n)}{n h(n)^{r+2d}} = \frac{0}{L^{2/p}}.$$
If $L \neq 0$ then this is sufficient to give a zero limit for the quantity of interest, which is what you were trying to establish.  If $L=0$ then you obtain an indeterminate form, and so the limit depends on the speed at which the numerator and denominator approach zero.  You have not specified enough information about the function $h$ to assess this.
So, the answer to your question is that the conditions you have are not sufficient to give the result you want, but if you are willing to add a couple more conditions then you can get this result.
A: Assume $h=h(n)$ is a function of the natural number $n$ that has a non-positive value only finitely often, so that the limits in the question are defined.  After squaring both sides of the right hand limit to remove the $1/2$ powers, rewrite the implication as
$$\lim_{n\to\infty}\frac{\log n}{n^{1-2/p}\,h(n)^r} \to 0\ \implies\ \lim_{n\to\infty}\frac{\log n}{n\,h(n)^{r+2d}} = 0.$$
The antecedent of this implication states $n^{1-2/p}\, h(n)^r$ grows faster than $\log(n);$ that is, for any natural number $M$ (no matter how large), eventually for all sufficiently large $n,$ $n^{1-2/p}h(n)^r \gt M\log(n).$  Its consequent says something similar.  Taking $n$ to be large enough for  both assertions, the translation is the following:

For any natural number $M,$ if it is the case that for sufficiently
  large $n,$ 
$$n^{1-2/p}h(n)^r \gt M\log(n),$$
then it is also the case that
$$n\,h(n)^{r+2d} \gt M\log(n).$$

Because eventually everything in sight is positive (that is, $n,$ $h(n),$ $\log n,$ $r,$ $d,$ and $p$), use the rules of algebra to rewrite this implication as
$$g(n) \gt 1\,\ \implies\ g(n) \gt \left(M\log n\right)^{1/(r+2d)-1/r}\, n^{(p-2)/(pr) - 1/(r+2d)}$$
with 
$$g(n) = h(n) \left(M\log(n)\right)^{-1/r}\,n^{(p-2)/(pr)}.$$
For this implication to hold, the right hand side of the consequent must (eventually, for all $M$ and sufficiently large $n$) not exceed the right hand side of the antecedent, equal to $1:$
$$\left(M\log n\right)^{1/(r+2d)-1/r}\, n^{(p-2)/(pr) - 1/(r+2d)}\le 1.$$
Since any power of $n,$ no matter how small, eventually exceeds any power of $\log(n),$ no matter how large, this implication can be true only when the power of $n$ is negative; that is,
$$\frac{p-2}{pr} \lt \frac{1}{r+2d}.$$
Since $p\gt 2$ and $r$ and $d$ are not negative (it doesn't matter whether they are integers), this can be a little more simply expressed by the relation
$$(p-2)(r+2d) \lt pr.$$
That, in turn, can be expressed as giving bounds for any one of $(p,r,d)$ in terms of the other two:

  
*
  
*$p \lt \frac{r}{d}+2;$
  
*$r \gt (p-2)d;$
  
*$d \lt \frac{r}{p-2}.$

