Question

Prove that the function g(x,y) = [(sin(x^2+y^2)/(sqrt(x^2+y^2)) if (x,y) does not = (0,0) ],
[0 if (x,y)=(0,0)]

is or is not continuous using the definition of continuity.

Answer 1

First what's your intuition tell you?

Answer 2

Well because of the piecewise specifically including 0 and the fact that 0,0 is not a solution I assume that its continuous. But i'm having trouble using the delta epsilon definition of continuity to prove it.

Answer 3

What do you know about this function
f(x) = sin x / x, x not = 0
= 1 , x = 0

Is it continuous?

Answer 4

yes because the limit as sin(x)/x goes to zero is 1

Answer 5

Right?

Answer 6

Yes. Now what about if instead of sin(x)/x, it were sin(x^2)/x ?

Answer 7

It has limit zero because
\[ \frac{\sin x^2}{x} = \frac{\sin x^2}{x^2} x = \left(\frac{ \sin x}{x}\right)^2 x \longrightarrow 0\]
as x --> 0

So yes, I think your intuition that the function you have also goes to zero as (x,y) --> 0 and therefore is continuous there.

Now to prove it. What kind of proofs are you using? Epsilon-delta?

Answer 8

sorry, that last bracketed expression is bogus, but even so it still goes to zero as
\[ \lim_{x \rightarrow 0} \frac{\sin x^2}{x^2} = 1 \]

Answer 9

...hence \[ \lim_{x \rightarrow 0} \frac{\sin x^2}{x^2} x = 1 \cdot 0 = 0 \]

Answer 10

Im using epsilon delta proofs.
I still don't quite understand them but I believe that I need to somehow get delta in terms of epsilon

Answer 11

Correct. Now notice that for x,y sufficiently small,
\[ | \sin(x^2 + y^2) | \leq x^2 + y^2 \]
That should give you the hint you need.

Answer 12

So the specific definition that i'm using states
If S is a metric space with metric d, ˆ S is a metric space with metric ˆ d,
and f : S → ˆ S, then f is continuous at p means that
(i) p ∈ S, and
(ii) if ǫ is a positive number, then there is a positive number δ such that if q is in S and
d(q, p) < δ, then ˆ d(f(q), f(p)) < ǫ.

so d(hat) (f(x,y),f(0,0))=If(x,y)-f(0,0)l=I(sin(x^2-y^2)/(sqrt(x^2+y^2))-0I ... following this point an example I have getting it to the form d(a,b)<epsilon but I don't know how to get there with this problem...

Answer 13

Well, notice that
\[ d(0,f(x,y)) = \left| \frac{\sin(x^2 + y^2)}{\sqrt{x^2 + y^2}} \right| \leq \frac{x^2 + y^2}{\sqrt{x^2 + y^2}} = \sqrt{x^2 + y^2} = d(0,(x,y)) \]

Now see what to do?

Answer 14

I see where all the pieces are coming from i just cant seem to straighten it out in my head.

so we have d(0,f(x,y)) and that needs to be <epsilon and d(0,(x,y)) which needs to be < delta

so sqrt(x^2+y^2) needs to be less than delta

and the function needs to be less than epsilon...

Answer 15

What you want to do is this: given any epsilon > 0, you want to find a delta (which is a function of epsilon) such that

d(0,(x,y)) < delta => d(0,f(x,y)) < epsilon

We have just shown above something incredibly useful. We have shown that
d(0,f(x,y)) < d(0,(x,y))

Hence given any such epsilon, there is an incredibly easy choice of delta such that
d(0,(x,y)) < delta => d(0,f(x,y)) < epsilon

Answer 16

so can we say delta=epsilon?

Answer 17

Yes. Given any epsilon > 0, choose delta = epsilon. Then
d((0,0),(x,y)) < delta = epsilon => d(0,f(x,y)) < epsilon
because
\[ d(0,f(x,y)) \leq d((0,0),(x,y)) \]

Answer 18

So that proves that the function is continuous everywhere because you can have a delta for any choice of epsilon?

Answer 19

No. It proves only that f is continuous at (0,0).

But the function is clearly continuous everywhere else, as it is the ratio of continuous functions, the denominator of which is not zero.

Answer 20

Oh, ok I forgot we were specifically using the point (0,0). Thanks a lot, that really helped to walk through it. We are just given the definition and not really taught how to use it.

Answer 21

ok. I'm sure in your textbook there are some worked examples and counter-examples. Go and read them carefully, even writing a couple of them out. Epsilon-delta proofs are very subtle and it takes a while to get used to them.

Answer 22

The text has them in a different format, that is what I tried first. Anyways ill keep practicing.

Answer 23

And don't feel bad about not "getting it" straight away. It's probably the most subtle kind of proof you've yet seen and lots of people take a while to get used to it.

Good luck.

Answer 24

Thanks!