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

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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).
 
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roam said:
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! :smile:

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? :confused:
 
Hi!

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

to prove it, use the definition of limit (y'know, the epsilon and delta thing) :wink:
 
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.
 
roam said:
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 :wink:
 
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.
 
tiny-tim said:
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 :wink:

You mean \epsilon = \frac{\delta}{2}?
 
No, he meant epsilon = 1/2.
 
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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!
 

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