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$$\lim_{r \to 0^+} \frac{\sqrt r}{(r-9)^4}\ $$

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?

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Did you mean $\displaystyle \lim_{r \to 0^+}$? –  user17762 Jun 18 '12 at 6:17
    
Whoops you are right –  soniccool Jun 18 '12 at 6:20
    
What does it approach if the limit approach 9 instead of 0 from the right? –  soniccool Jun 18 '12 at 6:26
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2 Answers

up vote 2 down vote accepted

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 $0$.

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How can i find which infinity it is approaching? –  soniccool Jun 18 '12 at 6:22
    
It is approaching $0$, as explained above. Not infinity of either kind. –  André Nicolas Jun 18 '12 at 6:24
    
What does it approach if the limit approach 9 instead of 0 from the right? –  soniccool Jun 18 '12 at 6:26
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Then the top will approach $3$. When $r$ is close to $9$, then $(r-9)^4$ will be very close to $0$ but positive (because of the $4$-th power). so when $r$ is close to $9$, we will be looking at something close to $3$ divided by a very small positive number. The result is a huge positive number, so the limit is $\infty$. –  André Nicolas Jun 18 '12 at 6:29
    
Okay i got it nothing fancy here its all just plugging in –  soniccool Jun 18 '12 at 6:31
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$\lim_{x \rightarrow 0^+} \frac{\sqrt{r}}{(r - 9)^4} = \frac{\sqrt{r}}{(r - 9)^4}$ 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.

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How can i find which infinity it is approaching? –  soniccool Jun 18 '12 at 6:21
    
@mystycs It is not approaching $\infty$. –  William Jun 18 '12 at 6:22
    
What does it approach if the limit approach 9 instead of 0 from the right? –  soniccool Jun 18 '12 at 6:26
    
@mystycs I have added the case when limit approaches $9$ to the answer. –  William Jun 18 '12 at 6:38
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