Limits of m/n as x Approaches 1

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The limit of the expression \( L = \lim_{x \to 1} \frac{x^m - 1}{x^n - 1} \) evaluates to \( \frac{m}{n} \) for natural numbers \( m \) and \( n \). The solution involves rewriting the limit using the formula for the sum of a geometric series, \( \sum_{k=0}^n x^k = \frac{x^{n+1} - 1}{x - 1} \). By factoring out \( (x - 1) \) from both the numerator and denominator, the limit simplifies to the ratio of the sums of constants, leading to the definitive conclusion that \( L = \frac{m}{n} \).

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lim xm-1/xn-1 m,n elements of N
x→1
the answer is m/n but i have no idea how to start or solve this!
 
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Consider the following:

$$\sum_{k=0}^n\left(x^k\right)=\frac{x^{n+1}-1}{x-1}$$

Can you now rewrite the limit to get a determinate form?
 
We are given to find:

$$L=\lim_{x\to1}\frac{x^m-1}{x^n-1}$$ where $$m,n\in\mathbb{N}$$

Using the hint I suggested, we may write:

$$x^m-1=(x-1)\sum_{k=0}^{m-1}\left(x^k\right)$$

$$x^n-1=(x-1)\sum_{k=0}^{n-1}\left(x^k\right)$$

And so our limit becomes:

$$L=\lim_{x\to1}\frac{(x-1)\sum\limits_{k=0}^{m-1}\left(x^k\right)}{(x-1)\sum\limits_{k=0}^{n-1}\left(x^k\right)}=\lim_{x\to1}\frac{\sum\limits_{k=0}^{m-1}\left(x^k\right)}{\sum\limits_{k=0}^{n-1}\left(x^k\right)}=\frac{\sum\limits_{k=0}^{m-1}\left(1\right)}{\sum\limits_{k=0}^{n-1}\left(1\right)}=\frac{\sum\limits_{k=1}^{m}\left(1\right)}{\sum\limits_{k=1}^{n}\left(1\right)}=\frac{m}{n}$$
 

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