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Find limit and find delta

  1. Sep 8, 2009 #1
    Let [tex] f(X)=\frac{\sin(2x)}{x} [/tex] and use a graphing utility to conjecture the value of L = [tex] \lim_{x->0}f(x) \mbox{ then let } \epsilon =.1 [/tex] and use the graphing utility and its trace feature to find a positive number [tex] \delta [/tex] such that [tex] |f(x)-l|< \epsilon \mbox{ if } 0 < |x| < \delta [/tex]. My conjecture of the limit L = 2, therefore if that is the case then [tex] 1.9< f(x) < 2.1[/tex]. Since the maximum value of f(x) < 2, the graphing utility will not be able to find delta will it? What is the value of delta if L=2?Thanks.
     
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  3. Sep 8, 2009 #2

    LeonhardEuler

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    It will be able to. The function does not need to go both above and below. You just need to find a number [tex]\delta[/tex] so that f(x) is between those numbers whenever x is in the interval [tex][-\delta,\delta][/tex].
     
  4. Sep 8, 2009 #3
    I set up my graphing calculator Xmin =.2758..., Xmax = .2775... Then f(x)=y=1.8999954 gives x_0=.27596197. The length 0 to x_0 is not = delta but this half interval has the property that for each x in the interval (except possibly for x=0) the values of f(x) is between either 0 and L+epsilon or L-epsilon. delta could be much smaller than x_0. I still have not found delta.
     
  5. Sep 8, 2009 #4

    LeonhardEuler

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    Any positive number delta that has that property ("for each x in the interval (except possibly for x=0) the values of f(x) is between either 0 and L+epsilon or L-epsilon") is a solution to the problem as long as it does not specifically ask for the largest possible number delta with that property.
     
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