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Thread: real hard problem

  1. #1
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    real hard problem

    Suppose that lim_{x\rightarrow a} f(x) exist and is positive. Show that for any positive integer n we have
    <br />
\lim_{x\rightarrow a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\rightarrow a} {f(x)}}<br />
.
    Last edited by mancillaj3; Apr 18th 2009 at 08:16 PM.
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  2. #2
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    Quote Originally Posted by mancillaj3 View Post
    Suppose that lim_{x\rightarrow a} f(x) exist and is positive. Show that for any positive integer n we have
    lim_{x\rightarrow a} \frac{n}{}\sqrt{f(x)} = \frac{n}{}\sqrt{lim_{x\rightarrow a} {f(x)}} .
    Those are meant to be n'th roots, right? \lim_{x\rightarrow a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\rightarrow a} {f(x)}} ?

    The slick way to do this is to say that \sqrt[n]{f(x)} is just the composition of the function f with the n'th root function g(t) = t^{1/n}. The function g(t) is continuous at all points t>0 (because in fact it is differentiable, with derivative \tfrac1n t^{(1/n)-1}). And the composition of two continuous functions is continuous. So the function \sqrt[n]{f(x)} is continuous at x=a (where f(a) is defined to be \lim_{x\to a}f(x)).
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  3. #3
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    thank you, is there another way to approach it?
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  4. #4
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    Quote Originally Posted by mancillaj3 View Post
    is there another way to approach it?
    You could use the identity s^n-t^n = (s-t)(s^{n-1} + s^{n-2}t + \ldots + st^{n-2} + t^{n-1}). This tells you that |s-t| = \frac{|s^n-t^n|}{s^{n-1} + s^{n-2}t + \ldots + st^{n-2} + t^{n-1}} \leqslant \frac{|s^n-t^n|}{t^{n-1}} provided that s and t are positive.

    Now suppose that \lim_{x\to a}f(x) = \ell>0. Put s = \sqrt[n]{f(x)} and t = \sqrt[n]\ell in the above inequality, to get \bigl|\sqrt[n]{f(x)} - \sqrt[n]\ell\bigr| \leqslant \ell^{(1-n)/n}|f(x)-\ell|. From that, you should be able to show that \lim_{x\to a}\sqrt[n]{f(x)} = \sqrt[n]\ell.
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