# 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.

$$\lim_{r\to 9} \frac {\sqrt{r}} {(r-9)^4}$$

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(You mean $r$, not $t$, right?) Are you sure that's the problem? It's of the form $3/0$ as written. – anon Feb 6 '12 at 0:28
do you mean "$r$ approaches 9"? – deinst Feb 6 '12 at 0:28

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.

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That line of thinking doesn't really make any sense to me at all. It isn't how things really work. For example doing that on this problem math.stackexchange.com/questions/106164/… causes a lot of problems. You would assume the answer is 2/0 which it is not. – user138246 Feb 6 '12 at 0:33
@Jordan: You're misreading the situation. We have $(-1)^3+1=0$, not $=2$ like you think. – anon Feb 6 '12 at 0:35
@Jordan, it doesn't work there, because you wind up with $0/0$ in that case (or "something small divided by something small", which does not lead you anywhere). Here everything is fine: the numerator is "more or less 3" and the denominator is small and positive (the "positive" is important). 3 divided by something small gets you something big... – David Mitra Feb 6 '12 at 0:39

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.

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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 $\infty$.

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