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Delta Epsilon Proof

  1. Nov 4, 2009 #1
    hey if lim (x-->0) f(x) = L

    where 0 < |x| < d1 implies |f(x) - L | < e

    how do i prove lim (x --> 0) f(ax) = L?

    i know

    0 < |ax| < |a|d1

    d2 = |a|d1

    but the textbook says d2 = d1/|a|

    help you guyssssssssssssssssssssssssssssssss
     
  2. jcsd
  3. Nov 4, 2009 #2

    Office_Shredder

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    You get to choose what d2 is
     
  4. Nov 5, 2009 #3
    You know that the function f is continuous, so you know that given epsilon > 0, there exists an delta > 0 satisfying
    d(x)<delta -> d(f(x),l))<epsilon. Now your problem asks you to show that, given an epsilon > 0, you can find a delta such that d(x)<delta -> d(f(ax),l)<epsilon.

    The point is that, you can also choose such a delta for ax. the continuity of the f immediately tells you that you can choose such a delta. What if f were the function f(x) = x^3? consider a neighborhood of f(x)=l=0, (-8,8). Then if x is between -2,2, we know that this relation holds. However, f(ax), a = 5, gives us a drastically stretched function. (-8, 8) is stretched to (-40, 40). if x is between -2/5, 2/5 = original delta/a, then d (f(ax), l) < d(f(x),l)<epsilon, because d(x) is now less than the first delta (2) in this case (because f(ax), x<delta over a implies that x from our second delta neighborhood, times a, is in our first delta neighborhood Adapt, and you'll be done.
     
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