# Homework Help: Limit log rule help

1. Feb 25, 2009

### KevinL

1. The problem statement, all variables and given/known data
Limit as p->infinity of ((p^2-p+1)/((P+1)^2)^(2p+3)

In case the parenthesis are confusing, its one giant fraction all raised to the (2p+3) power.

2. The attempt at a solution
I set the entire problem equal to L and took the ln of both sides. This lets me move the power down using a log rule So:

ln(L) = lim p->infinity (2p+3)* ln((p^2-p+1)/((P+1)^2)

Using the log rule of ln(m/n) = ln(m) – ln(n):

lim p-> infinity (2p+3)[ln(p^2-p+1) - 2ln(p+1)]

At this point Im not sure. I think I can put it into a form where I can then use hopital's rule? So:

[ln(p^2-p+1) - 2ln(p+1)] / (1/(2p+3))

Will that help? I took the derivative of top and bottom but its not looking like something I can use.

2. Feb 25, 2009

### Tom Mattson

Staff Emeritus
Re: limit

Let's clean this up a bit using LaTeX. If you want to see the code that I used to generate this image, just click on it. Make sure popups are allowed though.

$$\lim_{p\rightarrow\infty}\left(\frac{p^2-p+1}{(p+1)^2}\right)^{2p+3}$$

So far so good.

This isn't going to help too much because you get the indeterminate form $\infty - \infty$. Instead write it like this:

$$\ln(L)=\lim_{p\rightarrow\infty}(2p+3)\ln\left(\frac{p^2-p+1}{(p+1)^2}\right)$$

This is the indeterminate form $\infty \cdot 0$. Do you know how to handle that?

3. Feb 25, 2009

### KevinL

Re: limit

That latex code confuses me, but hopefully I can still write it out like ya would into a calculator.

So, with 0 * infinity I need to make it look like this:

ln((p^2-p+1)/((P+1)^2) / (1/(2p+3))

Ordinarily I would use hopital's rule, but wouldnt the top fraction become uglier after differentiating rather than more helpful?

4. Feb 26, 2009

### Tom Mattson

Staff Emeritus
Re: limit

It gets uglier before it gets better. If you take the derivative of the top and the bottom you end up with a ratio of rational expressions, which is itself a rational expression. Limits of rational expressions at infinity are easy to compute.