Suppose that $\displaystyle lim_{x\rightarrow a}$ $\displaystyle f(x)$ exist and is positive. Show that for any positive integer n we have

$\displaystyle

\lim_{x\rightarrow a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\rightarrow a} {f(x)}}

$.

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- Apr 16th 2009, 12:12 PMmancillaj3real hard problem
Suppose that $\displaystyle lim_{x\rightarrow a}$ $\displaystyle f(x)$ exist and is positive. Show that for any positive integer n we have

$\displaystyle

\lim_{x\rightarrow a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\rightarrow a} {f(x)}}

$. - Apr 17th 2009, 12:28 PMOpalg
Those are meant to be n'th roots, right? $\displaystyle \lim_{x\rightarrow a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\rightarrow a} {f(x)}}$ ?

The slick way to do this is to say that $\displaystyle \sqrt[n]{f(x)}$ is just the composition of the function f with the n'th root function $\displaystyle g(t) = t^{1/n}$. The function g(t) is continuous at all points t>0 (because in fact it is differentiable, with derivative $\displaystyle \tfrac1n t^{(1/n)-1}$). And the composition of two continuous functions is continuous. So the function $\displaystyle \sqrt[n]{f(x)}$ is continuous at $\displaystyle x=a$ (where $\displaystyle f(a)$ is defined to be $\displaystyle \lim_{x\to a}f(x)$). - Apr 18th 2009, 07:17 PMmancillaj3
thank you, is there another way to approach it?

- Apr 19th 2009, 07:45 AMOpalg
You could use the identity $\displaystyle s^n-t^n = (s-t)(s^{n-1} + s^{n-2}t + \ldots + st^{n-2} + t^{n-1})$. This tells you that $\displaystyle |s-t| = \frac{|s^n-t^n|}{s^{n-1} + s^{n-2}t + \ldots + st^{n-2} + t^{n-1}} \leqslant \frac{|s^n-t^n|}{t^{n-1}}$ provided that s and t are positive.

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