Finding limit of fraction with square roots: $\lim_{r\to 9} \frac {\sqrt{r}} {(r-9)^4}$ I have been looking at this for five minutes, no clue what to do.
$$\lim_{r\to 9} \frac {\sqrt{r}} {(r-9)^4}$$
 A: Hint: If $r$ is close to 9, then the numerator $\sqrt r$ is close to 3 and the denominator, $(r-9)^4$, is positive and close to 0. So, if you take a number close to 3 and divide by a small positive number, what do you get?  
If you take  numbers appoaching 3 and divide by   small positive numbers approaching 0, what do you get?  



Look at some specific values:
$$
\matrix{r& \quad\sqrt r\qquad&\qquad (r-9)^4\qquad & {\sqrt r\over (1-9)^4} \cr

9.1&\approx3 &.1^4 &\approx{3\over .1^4}=3\cdot 10,000  \cr

9.01&\approx3 &.01^4 &\approx{3\over .01^4}=3\cdot 10^8  \cr

9.001&\approx3 &.001^4 &\approx{3\over .001^4}=3\cdot 10^{12}  \cr


8.9&\approx3 &(-.1)^4 &\approx{3\over .1^4}=3\cdot 10,000  \cr

8.99&\approx3 &(-.01)^4 &\approx{3\over .01^4}=3\cdot 10^8  \cr

8.999&\approx3 &(-.001)^4 &\approx{3\over .001^4}=3\cdot 10^{12}  \cr



}
$$

Note the closer $r$ is to 9, the bigger $\sqrt r\over (r-9)^4$ becomes. So the limit is infinite.
Note also, please, that because the denominator is being raised to an even power, it is always positive.
The limit $\displaystyle\lim\limits_{r\rightarrow9} {\sqrt r\over (r-9)^3}$  is quite different, and in fact does not exist.
A: The limit is $+\infty$ because the numerator approaches a positive number and the denominator approaches $0$ from above.  Sometimes one says the limit "doesn't exist" when one means there is no real number that is the limit, so you could put it that way.
A: Replacing $r$ by $r-9$,
this becomes
$\lim_{r\to 0} \frac {\sqrt{r+9}} {r^4}
$. 
As was said, the numerator goes to 3
and the denominator goes to 0,
so the quotient goes to $\infty$.
