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.

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.
