Exhibit a series of functions on $(0,1)$ that converges to a continuous function but does not converge uniformly Let $f_n(x) = x^n$.  Prove ${f_n}$ converges on $(0,1)$ to a continuous function but does not converge uniformly.
Proof: Clearly, for any $\epsilon > 0$, for any $x$ in $(0,1)$, there exists $N$ such that for all $n > N$, $|f_n(x) - 0| < \epsilon.$  Therefore, $f_n$ converges to $f(x) = 0.$
For any $N > 0$, for all $\epsilon$ in $(0,1)$, there exists $x$ in $(0,1)$ such that $x> \epsilon^{\frac{1}{N+1}}.$  For such $x, |f_{N+1}(x)| > \epsilon.$  QED.
Is this proof correct? Is there a simpler way?
 A: I hope my reviews come across as constructive, I think that's what OP is looking for.

*

*The example is well chosen.

*If it is "clear" as it is written, why not prove it? Let $0<x_0<1$. Then $\frac{1}{x_0}=1+\epsilon $, with $\epsilon>0$. And $(\frac{1}{x_0})^n=(1+\epsilon)^n\geq1+n\epsilon$(by binomial expansion). So $\lim_{n \to +\infty} x_0^n$ exists and is equal to $0$. We then translate this as in the proposed proof where it is well done.

*It is well done except for one detail: it is better to write "towards the null function" or "to $f:(0,1)\to \mathbb R, x \mapsto 0$"than "to $f(x)=0".$

*Do not forget to notice that the null function is indeed continuous to conform to the statement of the question.

*For the second part of the proof, I would have written instead: let $n>0$. Let $0<(\frac12)^{1/n}<\xi<1$[which is possible because $(\frac12)^n=\exp(-\frac1nlog2)\to1^{-} $ when $n\to+\infty$]. Then $\xi^n>\frac12$. And we can conclude as OP.

To conclude, it is correct despite some flaws easily correctable by OP. This is a good example that can be nicely illustrated:

