I have been looking at this for five minutes, no clue what to do.
$\lim_{r\to 9} \frac {\sqrt{r}} {(r-9)^4}$
I have been looking at this for five minutes, no clue what to do.
$\lim_{r\to 9} \frac {\sqrt{r}} {(r-9)^4}$
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?
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.
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.
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 $\infty$.