Exploring Limits at (0,0) on a Parabola

In summary: Using an \epsilon- \delta proof, \epsilon is "given" and \delta depends on \epsilon, not the other way around.
  • #1
roam
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12

Homework Statement



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

a) Show that if C is any straight line through (0,0) then [tex]\lim_{(x,y) \to (0,0)}[/tex] 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 [tex]y \leq 0[/tex] and [tex]f(x,y) = 1[/tex]) but still I don't know how to "prove" part a).
 
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  • #2
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 [tex]y \leq 0[/tex] and [tex]f(x,y) = 1[/tex]) 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:
 
  • #3
Hi!

In part b) asks to 'show' that the limit as (x,y) -> 0 doesn't exist...
 
  • #4
Well, it obviously doesn't exist …

to prove it, use the definition of limit (y'know, the epsilon and delta thing) :wink:
 
  • #5
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.
 
  • #6
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:
 
  • #7
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.
 
  • #8
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 [tex]\epsilon = \frac{\delta}{2}[/tex]?
 
  • #10
Using an [itex]\epsilon- \delta[/itex] proof, [itex]\epsilon[/itex] is "given" and [itex]\delta[/itex] depends on [itex]\epsilon[/itex], not the other way around.

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

1. What does it mean to explore limits at (0,0) on a parabola?

Exploring limits at (0,0) on a parabola involves determining the behavior of the function as it approaches the point (0,0) on the graph of a parabola.

2. Why is (0,0) a special point on a parabola?

(0,0) is considered a special point on a parabola because it is the vertex of the parabola, where the direction of the curve changes from increasing to decreasing or vice versa.

3. How do you find the limit at (0,0) on a parabola?

To find the limit at (0,0) on a parabola, you can use the standard limit formula or evaluate the function at values approaching 0 from both sides of the axis. Alternatively, you can also use the concept of symmetry to determine the limit.

4. What is the significance of exploring limits at (0,0) on a parabola?

Exploring limits at (0,0) on a parabola can help us understand the behavior of the function at the vertex and provide insights into the overall shape and characteristics of the parabola. It is also a fundamental concept in calculus and is used in various applications.

5. Can the limit at (0,0) on a parabola be undefined?

No, the limit at (0,0) on a parabola will always exist as long as the function is defined at that point. The limit may approach positive or negative infinity, but it will not be undefined.

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