# Limits problem

1. May 26, 2012

### Saitama

1. The problem statement, all variables and given/known data
$$\stackrel{lim}{x→1}(\frac{p}{1-x^p}-\frac{q}{1-x^q})$$

2. Relevant equations

3. The attempt at a solution
I tried writing 1-xp as (1-x)(1+x2+x3........xp-1) and same with 1-xq but i don't seem to find any way further.

2. May 26, 2012

### I like Serena

Hi Pranav!

Have you tried turning it into 1 fraction first?

Since (1-x) is the factor that approaches zero, perhaps you can then divide that away in both nominator and denominator...

3. May 26, 2012

### Ray Vickson

Write $x = 1-r$ and ask for the limit as $r \rightarrow 0.$ Use the binomial expansions of $(1-r)^p, \; (1-r)^q$ in the denominators $1-x^p = 1-(1-r)^p \text{ and } 1-x^q = 1-(1-r)^q.$

RGV

4. May 27, 2012

### Saitama

Hello ILS!
At the first start, i did that but i stuck there too so i left it and tried to find a different way.

Hello RGV!
Even i try to expand, the denominators become zero again. If i solve 1-(1-r)^p, all the terms will have "r" with them and if i substitute r=0, the denominators again become zero.

5. May 27, 2012

### Infinitum

You need to combine both ILS and RGV's hints

The denominator does give 0, if you put in r=0. But, you can also take out r common from the denominator, and divide it in the numerator, which leaves you a constant term in denominator. Oh, and you need to apply binomial expansion even in the numerator.

$\frac{p(^qC_1r+^qC_2r^2+......)-q(^pC_1r+^pC_2r^2+......)}{r^2(^pC_1+^pC_2r+......)(^qC_1+^qC_2r+......)}$

6. May 27, 2012

### I like Serena

Well, you kind of have to.
As it is both fractions approach infinity.
So you are subtracting infinity from infinity, which is undetermined.

The only way I now to avoid that, is combining the 2 fractions into 1 fraction first.

Last edited: May 27, 2012
7. May 27, 2012

### Ray Vickson

Why do you want to put r = 0? You are being asked for a LIMIT as r → 0. That means that you need to find a value, say v, towards which the original expression tends as |r| becomes smaller and smaller and smaller ... . When we compute the limit we never need to put r = 0. In some cases a limit as r → 0 is the same value that would be obtained by actually putting r = 0, but in others (such as the current one in this thread) we cannot ever put r = 0.

RGV

8. May 27, 2012

### Ray Vickson

It is much easier to apply the binomial theorem first, to the two separate fractions, take out the common factor r from both denominators, and then combine the fractions. However, either way will lead to the same final answer, provided that one is careful.

RGV

9. May 27, 2012

### Saitama

Thank you all for the help!

Combining the hints provided by ILS and RGV i got my answer.
Here's the solution:
(I am dropping the limit word just to make it easier for me to write)

Substituting x=1-r as r→0,
$$\frac{p}{1-x^p}-\frac{q}{1-x^q}=\frac{p}{(^pC_1r-^pC_2r^2+......)}-\frac{q}{(^qC_1r-^qC_2r^2+......)}$$
$$=\frac{1}{r}[\frac{pq-p(^qC_2r-^qC_3r^2+.....)-pq+q(^pC_2r-^pC_2r^2+...)}{(^pC_1-^pC_2r+......)(^qC_1-^qC_2r+......)}]$$

pq cancels out. Taking out the common factor from numerator, and substituting r=0, i am left with,

$$\frac{^pC_2q-^qC_2p}{^pC_1 {}^qC_1}$$
Solving this, i get
$$\frac{p-1}{2}-\frac{q-1}{2}$$
Finally, i get the result
$$\frac{p-q}{2}$$

10. May 28, 2012

### Infinitum

Yep! That's perfect!