MHB Limits of m/n as x Approaches 1

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The limit of the expression as x approaches 1 is derived using the formula for the sum of a geometric series. The limit can be rewritten to eliminate indeterminate forms by factoring out (x-1) from both the numerator and denominator. This leads to the simplified limit expression, which evaluates to the ratio of the sums of constants, resulting in m/n. The calculations confirm that as x approaches 1, the limit converges to m/n, where m and n are natural numbers. This demonstrates the relationship between the powers of x in the limit.
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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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