Sequence $x^n$ doesn't converge uniformly to $f$ if $x \in[0,1]$

Question

I have a question about the proof of the next statement.

"The sequence of functions $x^n$ doesn't converge uniformly to some function $f$ if $x \in [0,1]$."

Here's the proof:

Let $0< \varepsilon <1$ and $x \in[0,1)$ such that $ \varepsilon^{1/n} \le x<1$.

In this case we have that $$f_n (x)-f(x)= \ x^n\ge \varepsilon \tag 1$$

Therefore$ \ x^n$ doesn't converge uniformly to some function $f$ if $x \in[0,1]$.

And I got stuck in part $(1)$, I mean why $f(x)=0$ ?

Answer 1

You have $x^n \to 0$ as $n\to\infty$ if $0\le x<1$ and $x^n\to1$ as $n\to\infty$ if $x=1.$

If a sequence of functions converges uniformly to a given function, then it also converges pointwise to the same function. In this case, a pointwise limit exists. So the remaining problem is to show that that pointwise limit is not a uniform limit. That is why $0$ is the thing to which uniform convergence would have to hold.