Pointwise and uniform convergence of fn(x) = (1-x)^(1/n)

[Vespero]

Homework Statement

Let fn(x) = (1-x)^(1/n) be defined for x an element of [0,1). Does the sequence {fn} converge pointwise? Does it converge uniformly?

Homework Equations

The formal definition of pointwise convergence:
Let D be a subset of R and let {fn} be a sequence of real valued functions
defined on D. Then {fn} converges pointwise to f if given any x in D and
given any epsilon > 0, there exists a natural number N = N(x, epsilon) such that
|fn(x) − f(x)| < epsilon for every n > N.

The formal definition of uniform convergence:
Let D be a subset of R and let {fn} be a sequence of real
valued functions defined on D. Then {fn} converges uniformly to f if given
any " > 0, there exists a natural number N = N(") such that
|fn(x) − f(x)| < epsilon for every n > N and for every x in D.

The Attempt at a Solution

I managed to prove that fn(x) converges pointwise by the following:
For x=0, fn(0) = (1-)^(1/n) = 1^(1/n) = 1, so lim(n->inf)fn(0) = 1.
For 0<x<1, 0 < (1 - x) < 1 and lim(n->inf)(1/n)=0, so lim(n->inf)fn(x) = (1-x)^0 = 1.
Thus, {fn(x)} converges pointwise to f(x) = 1.

However, I am having trouble proving whether or not it converges uniformly. By the definition, it converges uniformly if given and epsilon > 0, there exists a natural number N = N(epsilon) such that
|(1-x)^(1/n) - 1| < epsilon for ever n > N

I can see intuitively that for an epsilon, increasing n will cause (1-x)^(1/n) to approach one until the difference is less than epsilon, but how can I state this? Is there a way to explicitly set define N in terms of epsilon or vice versa? Is that necessary? Can I just explain that the first term of the difference approaches the second (and thus the difference approaches 0) as n increases, so there is an N for which the difference is less than epsilon when n > N?

 

[micromass]

I doubt very much that this sequence converges uniformly.
It is clear that this sequence does not converge uniformly on [0,1] (since the limit will not be continuous).

Maybe you can show the following: if the sequence converges uniformly on [0,1), then it will converge uniformly on [0,1]. I mean, you're just adding one point, that shouldn't mess with uniform convergence. But since the sequence does not converge uniformly on [0,1], it will then also not converge uniformly on [0,1)

Alternatively, you can take epsilon=1/2, and prove via the definition that the sequence does not converge uniformly...