Pointwise and Uniform Convergence of x^n where 0<x<1

[fluidistic]

Homework Statement

Almost like the title says : Let f_n (x)=x^n for 0 \leq x \leq 1. Find the pointwise limit of f_n and show that the convergence is not uniform.
What happens if x \in [0,1)?





2. The attempt at a solution
\lim _{n \to \infty} f_n = \lim _{n \to \infty} x^n = 0 if x is strictly lesser than 1 and it's worth 1 if x equals 1. (I wonder if I have to prove it via the definition of limits).
I don't have much ideas about the non uniform convergence. Checking up some wikipedia, I think that if I can show that if a_n does not tend to 0 when n tends to \infty, I'm done. Where a_n=sup |f_n (x)-f(x)|.
I get that a_n = sup |x(x^{n-1}-1)| but I think it's worth 0, hence implying the uniform convergence of f_n...


Answering the last question : By intuition the convergence is uniform.

 

[VeeEight]

The convergence is pointwise because the convergence of the sequence depends on what points of x you choose. Choosing x in [0,1), x=1, and x > 1 gives you a different limit.
You can use the definition of uniform convergence here: http://en.wikipedia.org/wiki/Uniform_convergence#Definition. If you are having trouble with pointwise vs. uniform convergence, compare the position of the phrase 'for all x in S' in the two definitions.

 

[fluidistic]

The convergence is pointwise because the convergence of the sequence depends on what points of x you choose. Choosing x in [0,1), x=1, and x > 1 gives you a different limit.
You can use the definition of uniform convergence here: http://en.wikipedia.org/wiki/Uniform_convergence#Definition. If you are having trouble with pointwise vs. uniform convergence, compare the position of the phrase 'for all x in S' in the two definitions.

Ok thanks, I'll try my best.