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Proving Limits

  • Thread starter roam
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  • #1
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Homework Statement



http://img40.imageshack.us/img40/39/20688555.gif [Broken]

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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Answers and Replies

  • #2
tiny-tim
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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
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Hi!

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

to prove it, use the definition of limit (y'know, the epsilon and delta thing) :wink:
 
  • #5
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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
tiny-tim
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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
HallsofIvy
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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
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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]?
 
  • #9
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No, he meant epsilon = 1/2.
 
  • #10
HallsofIvy
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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!
 

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