# Limit of x^n where 0<x<1

1. Oct 17, 2009

### fluidistic

1. The problem statement, all variables and given/known data
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.

2. Oct 17, 2009

### 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.

3. Oct 17, 2009

### fluidistic

Ok thanks, I'll try my best.