How Do Limits Behave at (0,0) on a Parabola?

[roam]

Homework Statement

http://img40.imageshack.us/img40/39/20688555.gif

a) Show that if C is any straight line through (0,0) then $$\lim_{(x,y) \to (0,0)}$$ along C exists and equals 1.

b) Show that the limit as (x,y) -> 0 doesn't exist.

Homework Equations

The Attempt at a Solution

I really need help with this question! I know that x² is a parabola and any straight line through the origin either intersects the parabola at some point and remains above it until 0 is reached (or lies on y=0, in which case $$y \leq 0$$ and $$f(x,y) = 1$$) but still I don't know how to "prove" part a).

 

[tiny-tim]

I know that x² is a parabola and any straight line through the origin either intersects the parabola at some point and remains above it until 0 is reached (or lies on y=0, in which case $$y \leq 0$$ and $$f(x,y) = 1$$) but still I don't know how to "prove" part a).

But you've done it!

If, sufficiently close to the origin, it remains above the parabola until 0 is reached, then the value along that part of the line is 1, so the limit exists …

what's worrying you about that?

 

[roam]

Hi!

In part b) asks to 'show' that the limit as (x,y) -> 0 doesn't exist...

 

[tiny-tim]

Well, it obviously doesn't exist …

to prove it, use the definition of limit (y'know, the epsilon and delta thing)

 

[roam]

Yes I see. I know how the epsilon & delta proof works but I don't know how to exactly apply it to this problem. I appreciate it if you could help me get started on this, thanks.

 

[tiny-tim]

Yes I see. I know how the epsilon & delta proof works but I don't know how to exactly apply it to this problem. I appreciate it if you could help me get started on this, thanks.

Well, if the limit exists, and is 1, say, then that means that given any epsilon > 0, we can find a ∂ > 0 such that √(x² + y²) < ∂ means that |f(x,y) - 1| < epsilon.

So ty it with epsilon = 1/2 ​

 

[HallsofIvy]

Simpler, I think: try approaching (0,0) along the curve y= (1/2)x². In that case, y is always less than x² so f(x,y)= 0.

 

[roam]

Well, if the limit exists, and is 1, say, then that means that given any epsilon > 0, we can find a ∂ > 0 such that √(x² + y²) < ∂ means that |f(x,y) - 1| < epsilon.

So ty it with epsilon = 1/2 ​

You mean $$\epsilon = \frac{\delta}{2}$$?

 

[Mark44]

No, he meant epsilon = 1/2.

 

[HallsofIvy]

Using an $\epsilon- \delta$ proof, $\epsilon$ is "given" and $\delta$ depends on $\epsilon$, not the other way around.

But I still think my suggestion is simpler. I'll just go off and sulk!