How can I use L'Hopital's rule to evaluate this exponential limit?

[f(x)]

Homework Statement

evaluate :
\lim_{x\longrightarrow1}\ \left\ ( \frac{p}{1-x^p} - \frac{q}{1-x^q} ) \ \right p,q\in N

2. What I've tried.

I s'pose L`Hopital's Cant be applied as this is not 0/0 form.
I took the LCM and put them in this form :

\lim_{x \longrightarrow 1}\ \frac{p-px^q -q+qx^p}{1-x^p-x^q+x^{p+q}}

But still i can't replace x-1 by h. How do i proceed
Thx

 

[MathematicalPhysicist]

take the exponential of this limit, and then youll have a limit with which you can calculate it via lhopital's rule.

 

[f(x)]

take the exponential of this limit, and then youll have a limit with which you can calculate it via lhopital's rule.

Sorry i don't understand what taking exponential means. Could you please explain ?
Thx

 

[MathematicalPhysicist]

well, calculate the limit of e^([p/(1-x^p)]-[q/(1-x^q)]), and use the fact that: lim e^g(x)=e^(lim (g(x))). (this is correct only if the limit is finite).

 

[f(x)]

well, calculate the limit of e^([p/(1-x^p)]-[q/(1-x^q)]), and use the fact that: lim e^g(x)=e^(lim (g(x))). (this is correct only if the limit is finite).

THx for the help, but how do i simplify the exponent? And since the denominator is constant, how do i apply lhopitals rule?

 

[Dick]

Try another approach. Since x->1, write x=1+y where y->0. Now write x^p as (1+y)^p and use the binomial expansion. Since y->0 you can ignore a lot of the higher powers in y. In this case you'll only need to keep the first three terms.

 

[f(x)]

Try another approach. Since x->1, write x=1+y where y->0. Now write x^p as (1+y)^p and use the binomial expansion. Since y->0 you can ignore a lot of the higher powers in y. In this case you'll only need to keep the first three terms.

Thx, that worked perfectly. I simplified and got (p-q)/2 which is the right answer.
But i have one question to ask, how did you decide to restrict to only 3 terms of the binomial expansion?

Once again, thanks a lot

 

[Dick]

Because I tried keeping just one term and it canceled exactly. Then realized I'd better keep another term. You can keep all of the terms if you want. But then you'll realize that the contributions beyond the third term go to zero as y->0.

 

[f(x)]

Because I tried keeping just one term and it canceled exactly. Then realized I'd better keep another term. You can keep all of the terms if you want. But then you'll realize that the contributions beyond the third term go to zero as y->0.

Yeah, fine.
Thank you very much