find this limit urgent!!!

Oct 2015
7
0
europe
limxm-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!
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
L'Hopital ...

$\displaystyle \lim_{x \to 1} \frac{x^m-1}{x^n-1} = \lim_{x \to 1} \frac{mx^{m-1}}{nx^{n-1}} = \frac{m}{n}\lim_{x \to 1} x^{m-n} = \frac{m}{n} \cdot 1 = \frac{m}{n}$

or, note ...

$x^m-1 = (x-1)(x^{m-1}+x^{m-2}+ ... + x^2 + x + 1)$

$x^n-1 = (x-1)(x^{n-1}+x^{n-2}+ ... + x^2 + x + 1)$

$\displaystyle \lim_{x \to 1} \frac{x^m-1}{x^n-1} =$

$\displaystyle \lim_{x \to 1} \frac{(x-1)(x^{m-1}+x^{m-2}+ ... + x^2 + x + 1)}{(x-1)(x^{n-1}+x^{n-2}+ ... + x^2 + x + 1)} =$

$\displaystyle \lim_{x \to 1} \frac{x^{m-1}+x^{m-2}+ ... + x^2 + x + 1}{x^{n-1}+x^{n-2}+ ... + x^2 + x + 1} = \frac{m \cdot 1}{n \cdot 1} = \frac{m}{n}$
 
Last edited:
May 2015
74
34
USA
It's not as elegant as Skeeter's method, but the binomial expansion makes the problem easier.

Since $x\to1$, define $x=1+h$, where $h$ represents the vicinity of $1$ and is going to 0 (i.e, $h\to0$) as $x\to1$.

$\begin{align}\frac{x^m-1}{x^n-1}&=\frac{(1+h)^m-1}{(1+h)^n-1}\qquad\text{When }h\to0\text{ the binomial expansion says it is approx. equal to }1+mh\text{ or }1+nh\cr
&=\frac{\cancel{1}+mh+\ldots\cancel{-1}}{\cancel{1}+nh+\ldots\cancel{-1}}\cr
&=\frac{mh+\ldots}{nh+\ldots}\cr
&\to\frac{m}{n}
\end{align}$