Question about specific step in proving Schur's Theorem (Combinatorics)

Question

I refer to p.98 of generatingfunctionology in proving Schur's Theorem:

The partial fraction expansion of $\mathcal{H}(x)$ is of the form

\begin{align*} \mathcal{H}(x) &= \frac{1}{(1-x^{a_1})(1-x^{a_2})\cdots(1-x^{a_M})} \\ &= \frac{c}{(1-x)^M} + O((1-x)^{-M+1}) \end{align*}

(where $a_1\le a_2\le \cdots\le a_M$ are M positive integers as defined earlier in the book.)

To calculate $c$, multiply both sides by $(1-x)^M$ and let $x\rightarrow 1$. This gives $c=\displaystyle\frac{1}{a_1\cdots a_M}$.

Question: can someone please help explain how multiplying both sides by $(1-x)^M$ and let $x\rightarrow 1$ would give $c=\displaystyle\frac{1}{a_1\cdots a_M}$? (I can see how the right hand side would reduce to $c$, but how does the left hand side reduce to $\displaystyle\frac{1}{a_1\cdots a_M}$?)

Answer 1

By L'Hopital's Rule, $$\lim_{x \to 1} \frac {1-x}{1-x^{a_k}}=\frac {1}{a_k}, \text{ so } \lim_{x \to 1} \prod_{k=1}^M \frac {1-x}{1-x^{a_k}}= \prod_{k=1}^M \frac{1}{a_k}.$$

Answer 2

For each factor of $1-x^{a_i}$ in the denominator, we have $$ \frac{1-x}{1-x^{a_i}} = \frac{1}{1 + x + x^2 + \cdots + x^{a_i-1}} $$ and as $x \to 1$, all $a_i$ of the terms in the denominator approach 1, so the overall fraction approaches $1/a_i$. Applying this limit to each term in the product $$ \frac{(1-x)^M}{\prod_{i=1}^M (1 - x^{a_i})} = \prod_{i=1}^M \frac{1-x}{1-x^{a_i}} $$ then yields the desired result.