Compute Limit Question Confused $$\lim_{r \to 0^+} \frac{\sqrt r}{(r-9)^4}\
  $$
How do i compute this limit? I was told to see what 1/x is approaching and rewrite it but can someone guide me in the right direction?
How can i find which infinity it is approaching?
Also
What does it approach if the limit approach 9 instead of 0 from the right?
 A: We need to look at what happens to top and to bottom as $r$ approaches $0$ from the right. The bottom behaves very nicely: as $r$ approaches $0$, $(r-9)^4$ approaches $(-9)^4$.
The top also behaves nicely: as $r$ approaches $0$ from the right, $\sqrt{r}$ approaches $0$.
So the quotient approaches $0/(-9)^4$, which is $0$.
Things get substantially more difficult when top and bottom both approach $0$. In that sort of situation, the analysis can be quite a bit harder to do. 
Remark: You might wish to confirm this with some calculator experimentation. Pick a very small positive $r$, like $r=10^{-6}$. Calculate $\frac{\sqrt{r}}{(r-9)^4}$. You will find it is close to $0$.
A: $\lim_{x \rightarrow 0^+} \frac{\sqrt{r}}{(r - 9)^4} = \frac{\sqrt{r}}{(r - 9)^4}$
since there is no $x$ occuring in the expression. 
However if you mean
$\lim_{r \rightarrow 0^+} \frac{\sqrt{r}}{(r - 9)^4} = \frac{0}{9^4} = 0$. To see this note that the numerator approaches $0$ from the right and the denominator approaches $(-9)^4 = 9^4$. You can also use a $\delta$-$\epsilon$ proof. 
If you want 
$\lim_{r \rightarrow 9^+} \frac{\sqrt{r}}{(r - 9)^4} = \lim_{\epsilon \rightarrow 0^+} \frac{\sqrt{9}}{\epsilon^4} = + \infty$. Note that the denominator is positive as approach from the right. In fact it is positive even if you approach to the left since, you are raising the denominator to the fourth power. 
