# Exploring Limits at (0,0) on a Parabola

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

## 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!

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 …

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)

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

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

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

No, he meant epsilon = 1/2.

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!

## 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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