MHB Limits of m/n as x Approaches 1

  • Thread starter Thread starter spartas
  • Start date Start date
  • Tags Tags
    Limits
Click For Summary
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.
spartas
Messages
7
Reaction score
0
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!
 
Mathematics news on Phys.org
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:

$$L=\lim_{x\to1}\frac{(x-1)\sum\limits_{k=0}^{m-1}\left(x^k\right)}{(x-1)\sum\limits_{k=0}^{n-1}\left(x^k\right)}=\lim_{x\to1}\frac{\sum\limits_{k=0}^{m-1}\left(x^k\right)}{\sum\limits_{k=0}^{n-1}\left(x^k\right)}=\frac{\sum\limits_{k=0}^{m-1}\left(1\right)}{\sum\limits_{k=0}^{n-1}\left(1\right)}=\frac{\sum\limits_{k=1}^{m}\left(1\right)}{\sum\limits_{k=1}^{n}\left(1\right)}=\frac{m}{n}$$
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
905
Replies
16
Views
890
  • · Replies 3 ·
Replies
3
Views
2K