Let $f_n(x)=x-x^n$ for $x \in [0,1]$ Does the sequence ${f_n}$ converge pointwise on the set $[0, 1]$?

Question

Let $f_n(x)=x-x^n$ for $x \in [0,1]$

Does the sequence ${f_n}$ converge pointwise on the set $[0, 1]$?

This is what I have done, since $0\le x\le 1$

$ x=0$$$f_n(0)=0-0^n=0$$ $x\to \infty$ $$f_n(x)=x-x^n=0$$$x=1$

$$f_n(1)=1-1^n=0$$

Therefore sequence pointwise converges to 0

Is this right?

Also how would I go about proving that $f_n$ is uniformly continuous on $[0,1]$

Answer 1

You shouldn't have $x\to\infty$, as $x\in [0,1]$.

For fixed $x$ you should ask yourself what $x^n$ does as $n$ gets large. There are two cases: $0\le x<1$ and $x=1$.

Answer 2

Note that for $0 < x < 1$, $0 < x^{n+1} < x^n$. Thus it is a monotone sequence which is bounded below, so it converges. Take the limit to get that $\lim_n x^n = 0$ for $0 < x < 1$ (I will leave it to you to prove this if you do not know this result already). Try out the other pieces by plugging in $x = 0$ and $x=1$ directly to see what you get.