1. ## Pointwise limit function

I have no notes on pointwise limit functions or their application to uniform convergence, and thus am having huge problems wth these questions because don't know how to start, or how to use my answer to determing uniform convergence.
For eachof the following sequences {fn} of functions, find the pointwise limit functon f (f it exists) on the given interval and decide whether {fn{ converges uniformly to f:
a) fn(x) = nthroot(x) on [0,1]
b) fn(x)=e^x / x^n on (1, infinity)
c) fn(x) = e^(-x^2)/n on R

Perhaps one worked through would be enough for me to tackle the other but I'm not sure please help!!!

2. Originally Posted by reeha
I have no notes on pointwise limit functions or their application to uniform convergence, and thus am having huge problems wth these questions because don't know how to start, or how to use my answer to determing uniform convergence.
For eachof the following sequences {fn} of functions, find the pointwise limit functon f (f it exists) on the given interval and decide whether {fn{ converges uniformly to f:
a) fn(x) = nthroot(x) on [0,1]
b) fn(x)=e^x / x^n on (1, infinity)
c) fn(x) = e^(-x^2)/n on R

Perhaps one worked through would be enough for me to tackle the other but I'm not sure please help!!!
If you are working with "uniform" convergence, you should be an old hand at "pointwise" convergence. That's the kind of convergence you work with in Calculus I and II.

For b), and c), it should be easy. Each is a fraction, the numerator is fixed and the denominator gets larger as n gets larger. The limit is obviously 0 for all x.

For a), look at an example. Suppose x= .6. Then, according to my calculator, $\sqrt{.6}= 0.77460$, $\sqrt[3]{.6}= 0.84343$, $\sqrt[4]{.6}= 0.88011$, $\sqrt[5]{.6}= 0.90288$, ..., $\sqrt[50]{.6}= 0.98983$, $\sqrt[100]{.6}= 0.99490$.

Can you guess what the limit will be? Can you see why that must be true?

To decide whether converge is uniform, think about how you would prove the pointwise limit. Given $\epsilon> 0$, you must find $\delta> 0$ such that... Given a specific $\epsilon$ does $\delta$ depend on the value of x or not? If no, then convergence is uniform.

3. I can see the limit is 1.

Well in lectures today we covered the M-test, so now I think I should probably use that. I still can't do it but I will have a go that way now. hmm.

4. Let's think about the first one.

You may recall that if we have a sequence of continuous functions $\{f_n\}$ that converges uniformly to $f$, then $f$ is continuous. We can first check where $\{\sqrt[n]{x}\}$ converges pointwise, on $[0,1]$.

If $x\ne 0$, then $\sqrt[n]{x}\to 1$. On the other hand, if $x=0$ then $\sqrt[n]{x}\to 0$. Is this result continuous? What can we conclude?