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Doing the limit we can see that in the open interval it converges pointwise to the constant function $f(x) = 0$. In the closed interval it doesn't converge uniformly because in $x=1$ $f(x) =1$ and when $0<x<1$ then $f(x)=0$. It isn't a continuous function although $f_n(x)$ is continuous for all $n$. So it can't be uniform.

However,let $0<ε<1$, we know the function increases as $x$ increases. So,

$|f_n(x) - f(x)| \leq ε^{n}$

And ε^(n) goes to 0 as n tends to infinity. Doesn't it mean that the convergence is uniform?

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2 Answers

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No, it doesn't mean that. Yes, $\lim_{n\to\infty}\varepsilon^n=0$, but that is not relevant here. Just note that$$(\forall n\in\mathbb N):\left(\sqrt[n]{\frac12}\right)^n=\frac12.$$But then $(f_n)_{n\in\mathbb N}$ doesn't converge uniformly to the null function. And, since it does converge pointwise to the null function, you can conclude that it doesn't converge uniformly at all.

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No, $x^n$ does not converge to zero uniformly on $[0,1)$, because

$$\lim_{n\to\infty}\sup_{0\leq x<1}|x^n|=1\neq 0$$

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