Why is $ \frac{\exp{\left( -\frac{x^2}{4y^{2r}} \right)}}{y^r} $ bounded on $[0,1]$? The following is a simple calculation I was trying to genearlize that I figured was just an application of l'hopitals rule.
Let $r >0$ 
1)  Why is $$ \frac{\exp{\left( -\frac{1}{4y^{2r}} \right)}}{y^r}  $$ bounded for $ y \in [0,1]$? 
2) What if I have a different power of $y$ in the denominator $ s>0$  what are the conditions I need on $r,s$ to guarantee $$ \frac{\exp{\left( -\frac{1}{4y^{2r}} \right)}}{y^s}  $$ is still bounded on $[0,1]$?
 A: We can do this for any $r,x>0$. Writing $t=1/y$, we get $$
\lim_{y\to0^+}\frac{\exp{\left( -\frac{1}{4y^{2r}} \right)}}{y^s}
=\lim_{t\to\infty}t^s\exp{\left( -\frac{t^{2r}}{4} \right)}
=\lim_{t\to\infty}\frac{t^s}{\exp{\left(\frac{t^{2r}}{4} \right)}}
=\lim_{t\to\infty}\frac{\exp{(s\log t)}}{\exp{\left(\frac{t^{2r}}{4} \right)}}
=\lim_{t\to\infty}\exp\left(s\log t-\frac{t^{2r}}{4}\right)=0,
$$
as $s\log t - t^{2r}/4\to-\infty$ as $t\to\infty$.
So the your function can be extended with continuity to zero. Thus we get a continuous function on the closed interval $[0,1]$ and so it is bounded.  
A: A variable change $\rm u=y^{-2r}$ puts expressions like this in the form
$$\rm u^k\exp(-au) ~~ \text{ as } ~~ u\to\infty.$$
Apply l'Hospital's $\rm \lceil k \rceil$ times (we assume $\rm k$ can be non-integer) and we obtain $0$.
We are assuming $\rm a,r>0$; otherwise we clearly have divergence.
A: Let $0<\delta<1$, then 
$$
f_{s,r}(y)=y^{-s}\exp\left(-\frac{1}{4y^{2r}}\right)
$$ 
is continuous on $[\delta,1]$. So it is bounded by some constant $M_1>0$. Now we note that substitution $y=(4t)^{-\frac{1}{2r}}$ where $t\to+\infty$ (recall $r>0$) gives us that
$$
\lim\limits_{y\to +0}f_{s,r}(y)=
\lim\limits_{y\to +0}y^{-s}\exp\left(-\frac{1}{4y^{2r}}\right)=
\lim\limits_{t\to +\infty}\frac{t^{\frac{s}{2r}}}{\exp(t)}=0
$$
The last limit is equal to $0$ because exponent grows faster then arbitrary power. Since $\lim\limits_{y\to +0}f_{s,r}(y)=0$, then $f_{s,r}(y)$ is bounded on $[0,\delta]$ by some $M_2>0$. Hence $f_{s,r}$ is bounded on $[0,1]$ by $\max(M_1,M_2)$
