Why is $\lim_{t\to 1}\frac{t^n-1}{t^m-1}=\frac{n}{m}$ $$\lim_{t\to 1}\frac{t^n-1}{t^m-1}=\lim_{t\to 1}\frac{(t-1)(t^{n-1}+t^{n-2}+...+t+1)}{(t-1)(t^{m-1}+t^{m-2}+...+t+1)}=\frac{n}{m}$$
How do you get $\frac{n}{m}$ in the end? 
 A: Hint
$$\underbrace{1^{n-1}+1^{n-2}+...+1+1}_{n\text{ addends}}=n$$
A: You can apply l'Hôpital's rule, we get that:
$$\lim_{\text{t}\to1}\frac{\text{t}^\text{n}-1}{\text{t}^\text{m}-1}=\lim_{\text{t}\to1}\frac{\text{nt}^\text{n-1}}{\text{mt}^\text{m-1}}=\lim_{\text{t}\to1}\frac{\text{n}\text{t}^{\text{n}-\text{m}}}{\text{m}}=\frac{\text{n}}{\text{m}}\lim_{\text{t}\to1}\text{t}^{\text{n}-\text{m}}=\frac{\text{n}}{\text{m}}\cdot1^{\text{n}-\text{m}}=\frac{\text{n}}{\text{m}}$$
A: Hint: $\underbrace{1 + 1 + \ldots + 1}_{n\text{ addends}}$ equals what?
Can you take it from here?
A: Because $$\frac{(t-1)(t^{n-1}+...+1)}{(t-1)(t^{m-1}+...+1)}=\frac{t^{n-1}+...+1}{t^{m-1}+...+1},$$
and all powers of $t$ converge to $1$ as $t$ goes to $1$. There are $n$ summands in numerator (the $n-1$, the $n-2$, ... the zeroth power of $t$) and $m$ summands in denominator. 
A: \begin{align*}
\lim_{t\to 1}\frac{t^n-1}{t^m-1}&=\lim_{t\to 1}\frac{(t-1)(t^{n-1}+t^{n-2}+...+t+1)}{(t-1)(t^{m-1}+t^{m-2}+...+t+1)}\\
&=\lim_{t\to 1}\frac{\overset{1\times n}{\overbrace{(t^{n-1}+t^{n-2}+...+t+t^{0})}}}{\underset{1\times m}{\underbrace{(t^{m-1}+t^{m-2}+...+t+t^{0})}}}\\
&=\frac{n}{m}
\end{align*}
A: $$\lim_{t\to 1}\frac{t^n-1}{t^m-1}=\lim_{t\to 1}\frac{(t-1)(t^{n-1}+t^{n-2}+...+t+1)}{(t-1)(t^{m-1}+t^{m-2}+...+t+1)}=\lim_{t\to 1}\frac{t^{n-1}+t^{n-2}+...+t+1}{t^{m-1}+t^{m-2}+...+t+1}$$
Substituting $t=1$:
$$\frac{\overbrace{1+1+1+...+1}^\text{n times}}{\underbrace{1+1+1+...+1}_\text{m times}}=\frac{n}{m}$$
