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Math Help - Nice limit

  1. #1
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    Nice limit

    find
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    Quote Originally Posted by dapore View Post
    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}.
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  3. #3
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    Quote Originally Posted by dapore View Post
    find
    You should know that:

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

    CB
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  4. #4
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    Thank you my friends
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  5. #5
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    how would you solve the problem if n,p were any non-zero real numbers? (of course L'Hospital's Rule is not allowed!)
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    Quote Originally Posted by NonCommAlg View Post
    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 ??
    Last edited by simplependulum; December 30th 2009 at 06:53 PM.
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    Quote Originally Posted by NonCommAlg View Post
    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...
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    Quote Originally Posted by Prove It View Post
    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 !!
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  9. #9
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    Couldn't you just perform long division?
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  10. #10
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    Quote Originally Posted by Prove It View Post
    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 ?
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  11. #11
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    Quote Originally Posted by Moo View Post
    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...
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  12. #12
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    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 ...
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  13. #13
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    Quote Originally Posted by NonCommAlg View Post
    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?
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