# Nice limit

• Dec 24th 2009, 10:56 PM
dapore
Nice limit
• Dec 24th 2009, 11:12 PM
Prove It
Quote:

Originally Posted by dapore

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 PM
CaptainBlack
Quote:

Originally Posted by dapore

You should know that:

$\displaystyle x^k-1=(x-1)(x^{k-1}+x^{k-2}+ ... + x^2+x+1)$

CB
• Dec 30th 2009, 01:23 AM
dapore
Thank you my friends
• Dec 30th 2009, 01:57 AM
NonCommAlg
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 PM
simplependulum
Quote:

Originally Posted by NonCommAlg
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)

This problem is equivalent to finding the derivative of $\displaystyle x^n$ from the first principle .

It is easy to solve when n is a rational number but how about irrational , or even transcendental ??
• Dec 31st 2009, 04:34 AM
Prove It
Quote:

Originally Posted by NonCommAlg
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)

Why can't you use L'Hospital? The numerator and denominator both tend to 0...
• Dec 31st 2009, 06:57 AM
simplependulum
Quote:

Originally Posted by Prove It
Why can't you use L'Hospital? The numerator and denominator both tend to 0...

It is a challenge he gives us , solving the limit without L'hospital , the difficuly rapidly increases !!
• Dec 31st 2009, 07:00 AM
Prove It
Couldn't you just perform long division?
• Dec 31st 2009, 07:03 AM
Moo
Quote:

Originally Posted by Prove It
Couldn't you just perform long division?

With n and p not being integers ?

But is it possible to use the 'definition' of the derivate number ?
• Dec 31st 2009, 07:06 AM
Prove It
Quote:

Originally Posted by Moo
With n and p not being integers ?

I don't see why not. The algebra could get quite messy though. I'm sure there's an easier way...
• Dec 31st 2009, 10:15 AM
NonCommAlg
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 AM
Drexel28
Quote:

Originally Posted by NonCommAlg
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 ...

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?