Now that the test is over...

[BobbyJoeCool]

I'm kindof wondering this... this is the question (Extra Credit)...

 

The following function is continuous at one and only one point. x=0. Prove it's continutity...

 

f(x)=0 when x is rational & fx^2 when x is irrational.

 

To prove a function is continuous at a point n...

 

1) f(n) must exist

2) [math]\lim_{x \to n}{f(x)}[/math] must exist.

3) [math]\lim_{x \to n}{f(x)}=f(n)[/math]

 

f(0)=0.

[math]\lim_{x \to 0}{f(x)}=0[/math] I have no idea how to prove this, but I know it's true because the question says that it's continous, and f(0) has to equal the limit, so the limit must be 0, but I don't know how to do this limit...

 

I know how to do the limit on a composition of functions, but this baffles me. anyone able to help?

[timo]

You are talking about f(x) = x² for x irrational, are you?

 

If I remember it correctly, convergation lim(x->0) f(x) = 0 demands that for all e>0, there is a d so that for all x in (-d, d) : |f(x)|<e.

Splitting up the reals into rationals and irrationals you see that this is trivially true for all rationals as 0<(e>0). For the irrationals in (-d, d) you have f(x) < d². Therefore you get the condition d²<e => d < sqrt(e). Since for all sqrt(e>0) there exists a d that satisfies this condition, your convergation criterion is fullfilled.

[Dave]

That's certainly the way I would do it. Epsilons and deltas are the way to go

[BobbyJoeCool]

isn't there another way to do it? He specifically told us that epsilon and delta proofs wouldn't EVER be used in this class... (then again, he assigned a problem on a quiz that required L'Hopital rule while still learning limits, as in no deriviatives)

 

The hint he gave us was the squeze theorm... I don't understand what it is from reading the book though...

 

and kx^2 for all x that are irrational doesn't have a limit (or a value) at x=0. Or so I was taught... because the graph is just a bunch of points, and it jumps around so much (because it also is the graph x=0 for rational) there is no limit at any number, but that this one is special and has a limit as x goes to 0, and is continious at x=0 (because the lines x=0, and kx^2 converge at 0)