# pointwise limit.

• Nov 10th 2010, 11:22 AM
Dreamer78692
pointwise limit.
if Fn(x) = x^(1/n) with Fn: [0,infinity) -> R

Find the pointwise limit of Fn(x)

this is what i have done:

x=0 Fn(x) -> 0
x!=0 Fn(x) -> 1

Is this correct... Im new to this so im not quite show about my answer

• Nov 10th 2010, 11:46 AM
Drexel28
Quote:

Originally Posted by Dreamer78692
if Fn(x) = x^(1/n) with Fn: [0,infinity) -> R

Find the pointwise limit of Fn(x)

this is what i have done:

x=0 Fn(x) -> 0
x!=0 Fn(x) -> 1

Is this correct... Im new to this so im not quite show about my answer

Ok, so how do you think we should show it (you're answer is correct)
• Nov 10th 2010, 12:06 PM
Dreamer78692
Do you mean how I arrived at my answer???
If so I used a theorem that states that a(1/n) -> 1 as n -> infinity where a is an element of R.

Sorry there is a second part to this question that is:

Does Fn converge uniformly on [0,1]

My answer is no since sup|x(1/n)| -> 1 as n->infinity

AND

Does Fn converge uniformly on [1/2,1]

My answer is the same as above

I seriously doubt this is correct

• Nov 10th 2010, 12:13 PM
Drexel28
Quote:

Originally Posted by Dreamer78692
Do you mean how I arrived at my answer???
If so I used a theorem that states that a(1/n) -> 1 as n -> infinity where a is an element of R.

Sorry there is a second part to this question that is:

Does Fn converge uniformly on [0,1]

My answer is no since sup|x(1/n)| -> 1 as n->infinity

AND

Does Fn converge uniformly on [1/2,1]

My answer is the same as above

I seriously doubt this is correct

Well, it's surely not uniformly convergent since each of the $f_n(x)$'s is continuous but the limit isn't (this is on $[0,\infty)$) That said, I would try re-evaluating the second question.