Find the limit $\lim_{x \to 1} \left(\frac{p}{1-x^p} - \frac{q}{1-x^q}\right) $ $p ,q >0$ I Know series expansion and L'Hospital's rule . But here both of them are not of any help.
 A: Let $$l=\lim_{x\rightarrow 1}\left(\frac{p}{1-x^p}-\frac{q}{1-x^q}\right)\;, p,q>0$$
Now put $\displaystyle x= \frac{1}{t}\;,$ Then $$l=\lim_{t\rightarrow 1}\left(-\frac{pt^p}{1-t^p}+\frac{qt^q}{1-t^q}\right) = \lim_{t\rightarrow 1}\left[\frac{p(1-t^p-1)}{1-t^p}-\frac{q(1-t^q-1)}{1-t^q}\right]$$
So $$l=(p-q)-l\Rightarrow l = \frac{p-q}{2}$$
A: $$x^m-1=(x-1)(x^{m-1}+x^{m-2}+\ldots+x+1)\implies$$
$$\lim_{x\to1}\frac p{1-x^p}-\frac q{1-x^q}=\lim_{x\to1}\frac{\frac p{x^{p-1}+\ldots+x+1}-\frac q{x^{q-1}+\ldots+x+1}}{1-x}=$$
$$\stackrel{\text{l'Hospital}}=-\lim_{x\to1}\left(-p\frac{(p-1)x^{p-2}+\ldots+1}{(x^{p-1}+\ldots+x+1)^2}+q\frac{(q-1)x^{q-2}+\ldots+1}{(x^{q-1}+\ldots+x+1)^2}\right)=$$
$$=\frac{p\frac{(p-1)p}2}{p^2}-\frac{q\frac{(q-1)q}2}{q^2}=\frac{p-1}2-\frac{q-1}2=\frac{p-q}2$$
A: If $p=q$ then the result will be $0$.
If $p \neq q$, then applying L'Hospital rule twice,
$\begin{align}
&\lim _{x\rightarrow 1}{\frac {p}{1-{x}^{p}}}-{\frac {q}{1-{x}^{q}}}\\
=&\lim_{x \to 1}{\frac {p-p{x}^{q}-q+q{x}^{p}}{ \left( -1+{x}^{p} \right)  \left( -1+{x}^{q} \right) }}\\
=&\lim_{x \to 1}{\frac {xpq \left( {x}^{q-1}-{x}^{p-1} \right) }{{x}^{p}p-{x}^{p+q}p+{x}^{q}q-{x}^{p+q}q}}\\
=&\lim_{x \to 1}{\frac {xpq \left( q{x}^{q-1}-p{x}^{p-1} \right) }{{x}^{p}{p}^{2}-{x}^{p+q}{p}^{2}-2\,{x}^{p+q}pq+{x}^{q}{q}^{2}-{x}^{p+q}{q}^{2}}}\\
=& \frac{p-q}{2}.
\end{align}$
A: Hint $\lim_{x \to 0} \frac{a^x-1}{x}=\ln(a)$ can you some how edit it to  use it to get  the resulting answer. 
A: Given limit can be set equivalent to by using $x=e^{t}$ and bit of simplification.
$$\lim_\limits{\large{t\to0}}\frac{1}{t}\left(\dfrac{(e^{pt}-1)}{tp}-\dfrac{e^{qt}-1}{tq}\right)$$
A: Here's a rather standard series expansion method: rewriting $x$ as $1+(x-1)$, and using Newton's generalised binomial expansion,
\begin{align}
  \frac{p}{1-x^p}-\frac{q}{1-x^q} & = \frac{p}{1-(1+(x-1))^p}-\frac{q}{1-(1+(x-1))^q} \\
                                  & = \frac{p}{1-(1+p(x-1)+\frac{p(p-1)}{2}(x-1)^2+O(|x-1|^3))} \\
                                  &-\frac{q}{1-(1+q(x-1)+\frac{q(q-1)}{2}(x-1)^2+O(|x-1|^3))} \\
                                  & = \frac{1}{(x-1)+\frac{q-1}{2}(x-1)^2+O(|x-1|^3)} \\
                                  &-\frac{1}{(x-1)+\frac{p-1}{2}(x-1)^2+O(|x-1|^3)} \\
                                  & = \frac{1}{x-1}(\frac{1}{1+\frac{q-1}2(x-1)+O(|x-1|^2)}-\frac{1}{1+\frac{p-1}2(x-1)+O(|x-1|^2)})\\
                                  & = \frac1{x-1}(\frac{p-1}2(x-1)-\frac{q-1}{2}(x-1)+O(|x-1|^2)) \\
                                  & = \frac{p-q}2.
\end{align}
Note that this requires no condition on $p,q$. They can even be negative.
