Limit and two numbers

dhiab · #1 $Old$ July 3rd, 2009, 09:48 AM

Fined this limit :

. n,p is the integers positifs numbers.

Prove It · #2 $Old$ July 3rd, 2009, 09:57 AM

Quote:

Originally Posted by dhiab $View Post$

Fined this limit :

. n,p is the integers positifs numbers.

You need to use the rule $a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2}b + a^{n - 3}b^2 + \dots + a^2b^{n - 3} + ab^{n - 2} + b^{n - 1})$

You should find something cancels and you're left with a nice easy sum...

dhiab · #3 $Old$ July 5th, 2009, 12:15 AM

Quote:

Originally Posted by Prove It $View Post$

You need to use the rule $a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2}b + a^{n - 3}b^2 + \dots + a^2b^{n - 3} + ab^{n - 2} + b^{n - 1})$

You should find something cancels and you're left with a nice easy sum...

Hello: you can let : x-1=t

Prove It · #4 $Old$ July 5th, 2009, 12:34 AM

Quote:

Originally Posted by dhiab $View Post$

Hello: you can let : x-1=t

Why would you want to do that? You're just bringing another variable into the problem.

Have you tried to use the method I told you?

Numerator:

$x^n - 1 = (x - 1)(x^{n - 1} + x^{n - 2} + x^{n - 3} + \dots + x + 1)$

Denominator:

$x^p - 1 = (x - 1)(x^{p - 1} + x^{p - 2} + x^{p - 3} + \dots + x + 1)$

So $\frac{x^n - 1}{x^p - 1} = \frac{(x - 1)(x^{n - 1} + x^{n - 2} + x^{n - 3} + \dots + x + 1)}{(x - 1)(x^{p - 1} + x^{p - 2} + x^{p - 3} + \dots + x + 1)}$

$= \frac{x^{n - 1} + x^{n - 2} + x^{n - 3} + \dots + x + 1}{x^{p - 1} + x^{p - 2} + x^{p - 3} + \dots + x + 1}$ .

So $\lim_{x \to 1}{\left(\frac{x^n - 1}{x^p - 1}\right)} = \frac{\lim_{x \to 1}({x^{n - 1} + x^{n - 2} + x^{n - 3} + \dots + x + 1})}{\lim_{x \to 1}({x^{p - 1} + x^{p - 2} + x^{p - 3} + \dots + x + 1})}$

Notice that there are exactly $n$ terms on top, and all of them will go to 1. There are exactly $p$ terms on the bottom and all of them will go to 1.

So $\lim_{x \to 1}{\left(\frac{x^n - 1}{x^p - 1}\right)} = \frac{n}{p}$ .