Is there a easy way to solve $\lim_{z \to 1} {{z^n -1} \over {z^m-1}}$ without L'Hospital?

Question

Is there a easy way to solve $$\lim_{z \to 1} {{z^n -1} \over {z^m-1}}$$ without L'Hospital?

Here $z \in \mathbb{C}$ and $m,n \in \mathbb{Z}$. Clearly $n/m $, using L'Hospital.

Answer 1

This is a difference of $n^\text{th}$ powers. You can write the fraction as: $$\begin{align*} \frac{z^n-1}{z^m-1}&=\frac{(z-1)\sum_{j=0}^{n-1}z^{n-j-1}}{(z-1)\sum_{j=0}^{m-1}z^{m-j-1}} \\ &=\frac{1+z+\cdots+z^{n-1}}{1+z+\cdots+z^{m-1}}, \end{align*}$$ using the partial sum of the geometric series: $\sum_{k=0}^{n-1}x^k=(x^n-1)/(x-1)$. This is now continuous at $z=1$, and so the limit is: $$\lim_{z\to1}\frac{z^n-1}{z^m-1}=\frac{1+1+\cdots+1\,(n\text{ times})}{1+1+\cdots+1\,(m\text{ times})}=\frac{n}{m}.$$