Continuity

**yarin012** · July 15th 2009, 12:34 PM

Hi everyone

I need to find whether this function is continues at (1,2).
Well actually the question is how do I find the limit as (x,y) goes to (1,2).
how do I do that?

Thank you

**arbolis** · July 15th 2009, 12:53 PM

In order for f to be continuous, $\lim _{(x,y) \to (1,2)}$ must be equal to $0$ .
First I notice that $\frac{\partial f}{\partial x}(1,2)$ and $\frac{\partial f}{\partial y}(1,2)$ doesn't seem to be easy to calculate. Hence we must use the definition of limit to prove what they ask for.
I'm not sure about how to proceed, but you have to reach $f(1,2)$ via a line and if it gives $0$ then try with a quadratic function. If it still gives $0$ , then you can suppose that the limit exists and is worth $0$ , hence the function to be continuous at $(1,2)$ . From it you would have to show that $\forall \varepsilon >0$ , $\exists \delta(\varepsilon)$ such that if $|(x,y)-(1,2)|<\delta(\varepsilon)$ , then $|f(x,y)-f(1,2)| < \varepsilon$ .
I'll wait for further help as well.

**yarin012** · July 15th 2009, 01:00 PM

You can't assume that the limit exists only by showing that it equals 0 via 2 different path. if you want to go that way you will have to prove that it holds for any curve of approaching.

**arbolis** · July 15th 2009, 01:05 PM

Originally Posted by yarin012

You can't assume that the limit exists only by showing that it equals 0 via 2 different path. if you want to go that way you will have to prove that it holds for any curve of approaching.

Yes I know. But this is not possible I believe. That's why you suppose it's true and try to prove it by definition. If you aren't able to prove it by any mean, then try another curve.
Edit: By the way, to show that it holds for any curve of approaching is precisely proving by definition that the limit exists!
Anyway I've plotted it and it didn't seem to be continuous. (I might be wrong of course). So you would have to try to approach the limit with a curve $y=ax+b$ and then $y=ax^2+bx+c$ . Hopefully it will be sufficient to prove that the limit doesn't exist.
Another edit : This is a hard one! I've tried to approach (1,2) by a curve y=ax, y=x^2, x=ay and it's all worth 0 if the constant a satisfy some equalities. Hence I've no idea how to show the limit doesn't exist.
I'd be glad if anyone could help us

**yarin012** · July 15th 2009, 02:05 PM

I know it's hard, I had to face it on a test.
However I know that you need to use the subs: u=(x-1) and v=(y-2).
and this facts:

1. $|uv|<=(u^2+v^2)/2$
2.

**arbolis** · July 15th 2009, 05:33 PM

Originally Posted by yarin012

I know it's hard, I had to face it on a test.
However I know that you need to use the subs: u=(x-1) and v=(y-2).
and this facts:

1. $|uv|<=(u^2+v^2)/2$
2.

Oh... thanks for the help. I almost get the result.
Here's my work: Let $u=(x-1)$ , $v=(y-2)$ . So the limit we're looking for (call it $I$ ) becomes equal to $\lim_{(u,v)\to (0,0)} \frac{uv^2}{u^2+\sin ^2(v)}$ .
Hence $I=\lim_{(u,v)\to (0,0)}\frac{|uv||v|}{u^2+\sin ^2 (v)} \leq \lim_{(u,v)\to (0,0)} \frac{(u^2+v^2)|v|}{2(u^2+ \sin ^2(v))}=\lim_{(u,v)\to (0,0)} \frac{|v|}{2}=0$ . As I $\geq 0$ , I must equal $0$ . Hmm this is only true if $u$ approaches $0$ from positive values.... So we're not yet done. I hope someone will finish it.

**yarin012** · July 15th 2009, 11:23 PM

A solution I got from a friend, I don't believe it's correct.

**At the last transition it should be (x-2) at the bottom instead of (x)

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