calculating the limit of a function without using L’Hopital’s rule

Question

Find the following limit without using L’Hopital’s rule:

$$\lim_{x\rightarrow 1} \frac{x^m-1}{x^n-1}$$

I think the limit using L'Hopital would be $\frac mn$ but I'm not sure how to find this without L'Hotpital's rule. TIA

Answer 1

We note that: $$ \sum_{k=0}^n x^k = \frac{x^{n+1}-1}{x-1} $$ As such, we have: $$ \frac{\sum_{k=0}^{m-1} x^k}{\sum_{k=0}^{n-1} x^k} = \frac{\left(\frac{x^{m}-1}{x-1}\right)}{\left(\frac{x^{n}-1}{x-1}\right)} = \frac{x^m-1}{x^n-1} $$ As such, $\lim_{x\to1} \frac{x^m-1}{x^n-1} = \lim_{x\to1} \frac{\sum_{k=0}^{m-1} x^k}{\sum_{k=0}^{n-1} x^k} = \frac{\sum_{k=0}^{m-1} 1^k}{\sum_{k=0}^{n-1} 1^k} = \boxed{\frac{m}{n}}$

Answer 2

You can use that $\sum\limits_{k=0}^{m-1} x^k=\frac{x^m-1}{x-1}$. Then you have

$$\lim_{x\to 1}\frac{x^m-1}{x^n-1}=\lim_{x\to 1}\frac{x-1}{x-1}\cdot \frac{\sum\limits_{k=0}^{m-1} x^k}{\sum\limits_{k=0}^{n-1} x^k}$$