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Help visualising a limit proof

  1. Jan 5, 2009 #1
  2. jcsd
  3. Jan 5, 2009 #2
    i cant see the animation of this proove
     
  4. Jan 5, 2009 #3

    HallsofIvy

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    When you don't know where to start, look at the definitions!

    "[itex]\lim_{x\rightarrow x_0} f(x)= L[/itex]" means, by definition, that

    "Given any [\itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]|x- x_0|< \delta[/itex] then [/itex]|f(x)- L|< \epsilon[/itex]".

    Here, the function is just f(x)= x and you want to prove that the limit is [itex]x_0[/itex]: write exactly the same thing but replace "f(x)" with "x" and "L" with "[itex]x_0[/itex]".

    "Given any [\itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]|x- x_0|< \delta[/itex] then [/itex]|x- x_0|< \epsilon[/itex]".

    You should see immediately that what was before "[itex]|f(x)- L|< \epsilon[/itex]" is now "[itex]|x- x_0|< \epsilon[/itex]" the same as with "[itex]|x- x_0|< \delta[/itex]". So make [itex]\epsilon[/itex] and [itex]\delta[/itex] the same: given any [itex]\delta> 0[/itex], you can always choose [itex]\delta= \epsilon[/itex].
     
  5. Jan 5, 2009 #4

    HallsofIvy

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    As for your picture, it would be better to draw it in two dimensions: draw the graph of y= x, a straight line. Since [/itex]\epsilon[/itex] is bounding the value of the function, draw two horizontal lines at [itex]Y= X+ \epsilon[/itex] and at [itex]y= X- \epsilon[/itex]. Draw vertical lines where the graph crosses those horizontal lines to get a "box" bounding the graph. The vertical lines will give [itex]X+\delta[/itex] and [itex]x-\delta[/itex]. Here, of course, because the line is y= x, that box is a square: [itex]\delta= \epsilon[/itex].
     
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