# Math Help - Nice limit

1. ## Nice limit

find

2. Originally Posted by dapore
find
Clearly the top and bottom both tend to 0 as $x \to 1$.

So you can use L'Hospital's Rule.

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

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

$= \frac{n}{p}$.

3. Originally Posted by dapore
find
You should know that:

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

CB

4. Thank you my friends

5. how would you solve the problem if $n,p$ were any non-zero real numbers? (of course L'Hospital's Rule is not allowed!)

6. Originally Posted by NonCommAlg
how would you solve the problem if $n,p$ were any non-zero real numbers? (of course L'Hospital's Rule is not allowed!)

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

It is easy to solve when n is a rational number but how about irrational , or even transcendental ??

7. Originally Posted by NonCommAlg
how would you solve the problem if $n,p$ were any non-zero real numbers? (of course L'Hospital's Rule is not allowed!)
Why can't you use L'Hospital? The numerator and denominator both tend to 0...

8. 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 !!

9. Couldn't you just perform long division?

10. 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 ?

11. 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...

12. here's a hint:
we only need to show that $\forall a \in \mathbb{R} : \ \lim_{x\to1} \frac{x^a - 1}{x-1} = a.$ it's easy to prove it for rational numbers. now for any $n \in \mathbb{N}$ choose $q_n \in \mathbb{Q}$ such that $q_n - \frac{1}{n} \leq a \leq q_n + \frac{1}{n}.$ then $\lim_{n\to\infty} q_n = a$ and ...

13. Originally Posted by NonCommAlg
here's a hint:
we only need to show that $\forall a \in \mathbb{R} : \ \lim_{x\to1} \frac{x^a - 1}{x-1} = a.$ it's easy to prove it for rational numbers. now for any $n \in \mathbb{N}$ choose $q_n \in \mathbb{Q}$ such that $q_n - \frac{1}{n} \leq a \leq q_n + \frac{1}{n}.$ then $\lim_{n\to\infty} q_n = a$ and ...
Yeah yeah, and then $\lim_{x\to1}\frac{x^a-1}{x^p-1}=\lim_{x\to1}\left\{\frac{x^a-1}{x-1}\cdot\frac{x-1}{x^p-1}\right\}=\frac{a}{p}$

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