Intuition behind f(x)=xnf(x)=x^n does not converge uniformly on (0,1)(0,1)

I would appreciate if someone could provide some intuition on why fn(x)=xnf_n(x) =x^n does not converge uniformly on (0,1)(0,1).

I am comfortable with the fact that fn(x)=xnf_n(x)=x^n does not converge uniformly on [0,1][0,1]. The limit function is not continuous. It seemed to me that the only point that was causing the trouble is x=1x=1. So I guessed that fnf_n would converge on (0,1)(0,1).

However, that was not the case, since we can always choose x=ϵ1/nx = \epsilon^{1/n}. Then we will have |xn|=|ϵ||x^n|=|\epsilon|. But it we move even further away from 11, namely (0,1−δ)(0,1-\delta), then it seems like we are good since |xn|<|(1−δ)n|→0|x^n|<|(1-\delta)^n|\to0 I can come up with the proof but it's a little bit counter intuitive to me 🙁 =================      When you are strictly away from the x=1x=1 then you do have uniform convergence. However, if you are arbitrarily close to x=1x=1, then your family of functions can't be uniformly convergent because it always have values very close to 11 for some fnf_n. – Jacky Chong Oct 21 at 1:22      I don't understand your "However, ...". xnx^n does indeed converge to 00 for any x∈(0,1)x\in (0,1). You can't "always choose x=ϵ1/nx = \epsilon^{1/n}", because that's not a fixed xx, it's dependent on nn. – Robert Israel Oct 21 at 1:25      @RobertIsrael How should I prove that xnx^n does not converge uniformly on (0,1)(0,1) then? – user34183 Oct 21 at 1:38      @RobertIsrael Also, why cannot xx be dependent on nn? I thought that's how you show not being uniformly convergent. Namely, it fails for some xx. – user34183 Oct 21 at 2:08      When you ask whether it converges on (0,1)(0,1), you need xx to be fixed. When you ask whether it converges uniformly, you can take xx dependent on nn. – Robert Israel 2 days ago ================= 2 Answers 2 ================= The intuition is that for x∈(0,1)x \in (0,1), xnx^n converges to 00: higher and higher powers of xx get smaller and smaller, and eventually less than any ϵ>0\epsilon > 0. But if xx is very close to 11, it takes a very high power to make xnx^n small, so the convergence is not uniform.

fn(x)≥max(0,1+n(x−1))f_n(x)\geq \max(0,1+n(x-1)) holds by convexity. For any x\in(0,1)x\in(0,1),
\lim_{n\to +\infty}f_n(x)=0\lim_{n\to +\infty}f_n(x)=0
hence the pointwise limit of \{f_n(x)\}_{n\geq 1}\{f_n(x)\}_{n\geq 1} is the zero function on (0,1)(0,1). Such limit cannot be a uniform limit, since
\forall n\geq 1,\qquad\sup_{x\in(0,1)} \left|f_n(x)-0\right|\geq \frac{1}{2} \forall n\geq 1,\qquad\sup_{x\in(0,1)} \left|f_n(x)-0\right|\geq \frac{1}{2}
by the initial inequality, if you like.