What is the Limit of a Fraction with Variable Exponents as x Approaches 1?

[Saitama]

Homework Statement

\stackrel{lim}{x→1}(\frac{p}{1-x^p}-\frac{q}{1-x^q})

Homework Equations

The Attempt at a Solution

I tried writing 1-x^p as (1-x)(1+x²+x³...x^p-1) and same with 1-x^q but i don't seem to find any way further.

 

[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...

 

[Ray Vickson]

Homework Statement

\stackrel{lim}{x→1}(\frac{p}{1-x^p}-\frac{q}{1-x^q})

Homework Equations

The Attempt at a Solution

I tried writing 1-x^p as (1-x)(1+x²+x³...x^p-1) and same with 1-x^q but i don't seem to find any way further.

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

 

[Saitama]

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...

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.

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

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.

 

[Infinitum]

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.

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+...)}

 

[I like Serena]

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.

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.

 

[Ray Vickson]

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.

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

 

[Ray Vickson]

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+...)}

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

 

[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}

 

[Infinitum]

Yep! That's perfect!