Real Analysis 2 question pointwise limits...?

Question

Could someone help me with this?? There is also a correction to the question: fn(0)=0

So when x= 0 fn(x)-->0
when x = 1/n² fn(x)-->+∞
when x=1/n fn(x)-->0

I don't know where it changes over from +∞ to 0 again. It's somewhere between 1/n² and 1/n but how do I figure that out??

Also if you can help with the other parts... that would be awesome!

http://i21.photobucket.com/albums/b273/q_midori/rahw2-4.jpg

The link is the entire question...

Answer 1

Actually, ∀x, f_n(x)→0 as n→∞. Why? Because if x is 0, then f_n(x) is always 0. Conversely, if x≠0, then ∃N∈ℕ s.t. 1/N < x. Then ∀n>N, we have that f_n(x) = 0. In either case, f_n(x) eventually reaches 0 and stays there, so the limit is 0, and the pointwise limit of f_n is simply the zero function.

Remember, in finding the pointwise limit of a function at a given point, you pick a point x, and keep it there - the point x is quite fixed, and does not change with n, so in particular, setting x=1/n² is not valid, because that point changes with n.

As for the other parts:

#2: In this case, ∀n, f_n is bounded, and is discontinuous at only one point (namely, x=0). Therefore, it is Riemann integrable. Further, it is equal almost everywhere to the continuous function g_n(x) = {n-n²x if 0≤x≤1/n, 0 if 1/n≤x≤1} (indeed, it differs from it at only the point 0), so its integral is the same, and thus [0, 1]∫f_n = [0, 1]∫g_n = [0, 1/n]∫n-n²x dx + [1/n, 1]∫0 dx = nx-n²x²/2 |[0, 1/n] + 0 = n(1/n) - 1/2 n²(1/n)² = 1-1/2 = 1/2. This holds for all n, so we have that [n→∞]lim [0, 1]∫f_n = [n→∞]lim 1/2 = 1/2.

#3: Since the pointwise limit of the f_n is just the 0 function, we have that [0, 1]∫f = 0 ≠ 1/2.