find http://mathramz.com/xyz/latexrender/...8e79587aba.png

Printable View

- Dec 24th 2009, 10:56 PMdaporeNice limit
- Dec 24th 2009, 11:12 PMProve It
Clearly the top and bottom both tend to 0 as $\displaystyle x \to 1$.

So you can use L'Hospital's Rule.

$\displaystyle \lim_{x \to 1}\frac{x^n - 1}{x^p - 1} = \lim_{x \to 1}\frac{\frac{d}{dx}(x^n - 1)}{\frac{d}{dx}(x^p - 1)}$

$\displaystyle = \lim_{x \to 1}\frac{nx^{n - 1}}{px^{p - 1}}$

$\displaystyle = \frac{n}{p}$. - Dec 24th 2009, 11:39 PMCaptainBlack
- Dec 30th 2009, 01:23 AMdapore
Thank you my friends

- Dec 30th 2009, 01:57 AMNonCommAlg
how would you solve the problem if $\displaystyle n,p$ were any non-zero

__real__numbers? (of course L'Hospital's Rule is not allowed!) (Evilgrin) - Dec 30th 2009, 06:17 PMsimplependulum
- Dec 31st 2009, 04:34 AMProve It
- Dec 31st 2009, 06:57 AMsimplependulum
- Dec 31st 2009, 07:00 AMProve It
Couldn't you just perform long division?

- Dec 31st 2009, 07:03 AMMoo
- Dec 31st 2009, 07:06 AMProve It
- Dec 31st 2009, 10:15 AMNonCommAlg
here's a hint:

we only need to show that $\displaystyle \forall a \in \mathbb{R} : \ \lim_{x\to1} \frac{x^a - 1}{x-1} = a.$ it's easy to prove it for rational numbers. now for any $\displaystyle n \in \mathbb{N}$ choose $\displaystyle q_n \in \mathbb{Q}$ such that $\displaystyle q_n - \frac{1}{n} \leq a \leq q_n + \frac{1}{n}.$ then $\displaystyle \lim_{n\to\infty} q_n = a$ and ... - Jan 1st 2010, 11:40 AMDrexel28
Yeah yeah, and then $\displaystyle \lim_{x\to1}\frac{x^a-1}{x^p-1}=\lim_{x\to1}\left\{\frac{x^a-1}{x-1}\cdot\frac{x-1}{x^p-1}\right\}=\frac{a}{p}$

By the way, was your point that if we can prove that the above limit is true for any $\displaystyle a\in\mathbb{Q}$ and since any real number has a sequence of rational points which converges to it we are done?