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Consider the sequence of functions $(f_n)_n$ given by $$ f_n : [0,1) \rightarrow \mathbb{R}: x \mapsto x^n. $$ I need to determine if this sequence converges pointwise and/or uniform.

First, I think it converges pointwise to the zero function, since $\lim_{n \to \infty} f_n = 0$ for every $x \in [0, 1)$. I could not complete my proof though.

Let $x \in [0, 1)$ be arbitrary. Let $\epsilon > 0$. Then we have to find a $n_0 \in \mathbb{N}$ such that from $\forall n \geq n_0$ it follows that $|f_n(x) - 0| < \epsilon$. Now we have $$|f_n(x) - 0| = x^n $$ Should I now consider the separate cases: $x = 0$ and $x \neq 0$? If we let $x \neq 0$ first, then we would have $x^n < \epsilon $ if $n \ln(x) < \ln(\epsilon)$, i.e. if $n > \ln(\epsilon) / \ln(x)$. Then I would choose $$ n_0 > \ln(\epsilon) / \ln(x). $$The case $x = 0$ is trivial.

For uniform convergence, I have no idea. But is the above approach correct?

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We have \begin{equation} \sup_{x\in [0,1)}|f_n(x)| = 1, \end{equation} which is not going to 0. Hence there's no uniform convergence. See

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Theorem: Let $\{f_n\}_n$ be a sequence of function defined on $E$ and for every $x\in E$, let $\lim_nf_n(x)=f(x)$. Let $$M_n=\sup_{x\in E}|f_n-f|$$ Then $f_n\to f$ uniformly on $E$ if and only if $\lim_{n\to \infty} M_n=0$

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