Why $f_n(x)=\sqrt[n]{x}$ on $[0,1]$ doesn't converge uniformly?

I was solving this problem:

For the following {$$f_n$$} sequence, determine the pointwise limit of {$$f_n$$} (if it exists) on the invterval, and indicate if {$$f_n$$} converges uniformly towards this function.$$f_n(x)=\sqrt[n]{x}, on [0,1]$$

I ended up with the pointwise limit being $$\lim_{n\to\infty}x^{1/n}=1$$ if $$0 and $$\lim_{n\to\infty}x^{1/n}=0$$ if $$x=0$$. My problem is to determine if it does or doesn't converge uniformly, I saw a question about it on this website, but I didn't understand it. The answer was that it doesn't converge uniformly, can you explain me why? Thanks.

• read through the definition of uniform convergence – Sean Nemetz Feb 9 at 23:34
• What is the limit of $f_n(x)$ as $n\rightarrow \infty$? Is it continuous on [0,1]? – Theo C. Feb 9 at 23:36
• @TheoC. it's not helpful, because that's (continuity of limiting function) not the definition of uniform convergence (but merely a consequence). – enedil Feb 9 at 23:37
• The limit is $0$ if $x=0$ and $1$ if $0<x\leq 1$, so it isn't continous on $[0,1]$. – davidllerenav Feb 9 at 23:38
• Show we can't find $N$ that does not depend on $x$ such that $|f_n(x) - f(x) | < \epsilon$ for $n > N$ and for ALL $x \in (0,1]$. – RRL Feb 9 at 23:51

$$|1 -f_n(2^{-n})| = \frac{1}{2} \not\to 0$$
• I don't understand what you did there. Where does that $f_n(2^{-n})$ comes from? – davidllerenav Feb 10 at 0:13
• The limit is $f(x) = 1$ for $0 < x \leqslant 1$ and $f(0) = 0$. For uniform convergence we must have for any $\epsilon > 0$, $N(\epsilon)$ such that $|f_n(x) - f(x)|< \epsilon$ for all $n > N$ and for all $x \in (0,1]$. Take $\epsilon = 1/4$ say. So if $x \neq 0$ we have $|f_n(x) - f(x)| = 1 - x^{1/n}$ But for any $n$ no matter how large if we look at $x = 2^{-n}$ we get $1 - (2^{-n})^{1/n} = 1/2$ and this is not less than $1/4$. – RRL Feb 10 at 0:21
• I see. You choose $x=2^{-n}$ for convenience, right? In the case that $x=0$, it woudl be $|0-x^{1/n}| right? – davidllerenav Feb 10 at 0:31 • Yes it was a convenient choice. We don't need to worry about$0$where convergence is automatic since$f_n(0) = f(0) = 0$. It is points near$0$in$(0,1]$that cause problems. If$f_n$converges uniformly on$[0,1]$then it converges uniformly on$(0,1]$. So if it fails to converge uniformly on$(0,1]$that is enough to show. – RRL Feb 10 at 0:34 • Ok, thanks. What other$x\$ could I choose? – davidllerenav Feb 10 at 1:22