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Given: $$f_n(x) = \sqrt[n]{x} \; , x \in [0,1]$$

Is the above series of functions converge uniformly? and how do we check it?

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No. You could start by graphing the first few functions to get an idea of what is going on. – Umberto P. Apr 24 '13 at 16:06
up vote 1 down vote accepted

One more way of seeing it:

Take $x \in (0,1]$. Then $\lim_{n \to \infty} x^{\frac{1}{n}}=1.$ Now take $x=0$. $\lim_{n \to \infty}x^{\frac{1}{n}}=0 \ne 1$. Hence the function does not converge uniformly, but it converges uniformly for $x \in (0,1]$

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How is $\lim_{n \to +\infty} x^{\frac{1}{n}} = 0$ ? – TheNotMe Apr 24 '13 at 16:30
Pls see the edit – Alex Apr 24 '13 at 16:34
Thanks. So the reason is that each $f_n{x}$ is continuous but $f(x)$ is not ? – TheNotMe Apr 24 '13 at 16:38
Yes. Have a look at this video lecture: – Alex Apr 24 '13 at 16:43
It doesn't converge uniformly in $(0,1]$ (consider for each $n$ the point $x_n=(\frac12)^n$), but it does converge to a continuous function, if that's what you meant. – Javier Apr 24 '13 at 17:44


$$\sqrt[n] a\xrightarrow[n\to\infty]{}1\;,\;\;\forall\,\,0<a\in\Bbb R$$

$$\sqrt[n] 0=0$$

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Ah. So it is because each $f_n(x)$ is continuous but $f(x)$ is not? – TheNotMe Apr 24 '13 at 16:17
Well, yes: that's basically the bottom line in the answers and comments... – DonAntonio Apr 24 '13 at 18:26

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