Show the limit is zero without doing integration Show that $$\lim _{t \to 0^ + } \frac{C}{t^{n/2}} \int_\delta^\infty e^{-r^2/(16t)} r^{n - 1} \,dr = 0$$ where $C,\delta$ are constants.
I believe there is a way to show the limit is zero without doing the integration. Hope someone can help. Thank you!
 A: Note that for $n\ge 1$ $$\lim_{t\to 0+}\frac{e^{-r^2/{16t}}}{t^{n/2}}=\exp\left(\lim_{t\to 0+}(-r^2/{16t}-n/2\ln t)\right)$$ Now note that the function $$f(t)=r^2/{16t}+n/2\ln t$$ attains its minima at $t=\frac{r^2}{8n}$ and hence as $t<r^2/8n$  $\forall \epsilon >0$, $\exists \delta>0$ such that $0<t<\delta\implies$ $f(t)>\epsilon\implies \lim_{t\to 0+} f(t)=\infty\implies $ $$\lim_{t\to 0+}\frac{e^{-r^2/{16t}}}{t^{n/2}}=0$$ Now apply monotone convergence theorem to show the desired result. 
A: Change variables using $u = \frac{r}{4\sqrt{t}}$, then $r = 4\sqrt{t} u$ and your expression becomes
$$\frac{C}{t^{n/2}}\int_{\frac{\delta}{4\sqrt{t}}}^\infty e^{-u^2} u^{n-1} (4\sqrt{t})^{n-1} * 4\sqrt{t} du$$
Now we can remove $t$ from the integral. Doing this carefully, I get:
$$\frac{(4\sqrt{t})^nC}{t^{n/2}}\int_{\frac{\delta}{4\sqrt{t}}}^\infty e^{-u^2} u^{n-1} du = 4^n C\int_{\frac{\delta}{4\sqrt{t}}}^\infty e^{-u^2} u^{n-1} du $$
Which appears to be independent of $t$ except for the lower limit of integration. Do you see where to take it from here?
