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:

$$L=\lim_{x\to1}\frac{(x-1)\sum\limits_{k=0}^{m-1}\left(x^k\right)}{(x-1)\sum\limits_{k=0}^{n-1}\left(x^k\right)}=\lim_{x\to1}\frac{\sum\limits_{k=0}^{m-1}\left(x^k\right)}{\sum\limits_{k=0}^{n-1}\left(x^k\right)}=\frac{\sum\limits_{k=0}^{m-1}\left(1\right)}{\sum\limits_{k=0}^{n-1}\left(1\right)}=\frac{\sum\limits_{k=1}^{m}\left(1\right)}{\sum\limits_{k=1}^{n}\left(1\right)}=\frac{m}{n}$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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