Limit Help: lim {x->1} {(m/(1-x^m))-(n/(1-x^n))}

[Messagehelp]

This limit has been killing me for two weeks now, and no matter how I try, I can't calculate the result algebraically.

$\displaystyle \lim_{x\to{1}}{\frac{m}{1-x^m}-\frac{n}{1-x^n}}$

Where m, n: arbitrary positive integers.

By manual testing (well semi-manual, by writing a script to do it), I found the result to be (m-n)/2, but no matter what I do on paper it always remains an indeterminate form. My only lead is the rule:

$\displaystyle \lim_{x\to{a}}{\frac{x^n-a^n}{x-a}} = n{a^{n-1}}$

Applying it does simplify the limit, but it still remains a 0/0 or inf-inf.

Any help or at least some pointers would be VERY welcome.

 

[o_O]

Do you know l'hopital's rule? If you apply it a couple of times after you combine the two fractions into one, you should arrive at the right answer:

$\displaystyle \lim_{x \to 1} \left(\frac{m}{1 - x^{m}} - \frac{n}{1-x^{n}}\right)$

$\displaystyle = \lim_{x \to 1} \left( \frac{m}{1 - x^{m}} \cdot \left(\frac{1 -x^{n}}{1-x^{n}}\right) \: - \: \frac{n}{1 - x^{n}} \cdot \left(\frac{1-x^{m}}{1-x^{m}}\right)\right)$

$\displaystyle = \lim_{x \to 1} \left( \frac{m - mx^{n} - n + nx^{m}}{(1-x^{m})(1-x^{n})}\right)$

\(\displaystyle = \lim_{x \to 1} \left( \frac{m - mx^{n} - n + nx^{m}}{1 - x^{n} - x^{m} + x^{m+n}} \right) = \left[\frac{0}{0}\right] \quad \mbox{Apply l'hopital's rule here}\)

Excuse me if I made some errors. But the idea is there. Combine into one fraction, apply l'hopital, simplify, apply l'hopital, simplify ....

 

[galactus]

Double O"s way is how I approached it. Was going to post earlier but didn't have time.

$\displaystyle \lim_{x\to{1}}\left[\frac{m}{1-x^{m}}-\frac{n}{1-x^{n}}\right]$

$\displaystyle =\lim_{x\to{1}}\frac{m-mx^{n}-n+nx^{m}}{(1-x^{m})(1-x^{n})}$

Now, use L'Hopital if you wish and get:

$\displaystyle \lim_{x\to{1}}\frac{mn(x^{m}-x^{n})}{((m+n)x^{n}-m)x^{m}-nx^{n}}$

Use L'Hopital again:

$\displaystyle \lim_{x\to{1}}\frac{mn(mx^{m}-nx^{n})}{((m+n)^{2}x^{n}-m^{2})x^{m}-n^{2}x^{n}}$

Now, if you look close, as x approaches 1 we have:

$\displaystyle \frac{mn(m-n)}{(m+n)^{2}-m^{2}-n^{2}}$

Which, when doing the algebra, reduces to:

$\displaystyle \frac{m-n}{2}$

 

[Messagehelp]

I love you, people! I've tried L'Hopital's rule but got stuck with the x^(n-1) and x^(m-1) parts. Now that I see the simplifications it seems so obvious! My sincerest gratitude for your help.