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Homework Help: Finding Delta Algebraically Thomas' Calculus

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data

    This is Example 5 in Chapter 2.3 of the above mentioned text:


    Prove that the [tex]\lim_{x\rightarrow2}f(x)=4[/tex] if [tex]f(x)= x^2 \text{ for }x\ne2\text{ and }f(x)=1\text{ for }x=2[/tex]


    Step 1 Solve the inequality [tex]|f(x)-4|<\epsilon[/tex] to find an open interval containing xo = 2 on which the inequality holds for all [itex]x\ne x_0[/itex]

    For [itex]x\ne x_0 = 2[/itex], we have f(x) = x2 and the inequality to solve is

    Thus, (Author's comments in Blue) :


    [tex]-\epsilon < x^2-4<\epsilon[/tex]

    [tex]4-\epsilon < x^2 < 4 + \epsilon[/tex]

    [tex]\sqrt{4-\epsilon} < |x| < \sqrt{4 + \epsilon}[/tex] Assumes [itex]\epsilon < 4[/itex]

    [tex]\sqrt{4-\epsilon} < x < \sqrt{4 + \epsilon}[/tex] An open interval about xo=2 that solves the inequality.

    I will stop here, though I have more questions. My initial 2 questions are these:

    1.) Maybe it's obvious, but in the last step, how did he drop the abs value sign?


    2.) I fail to understand the last blue comment. How does the preceding procedure take into account that xo = 2 ?

  2. jcsd
  3. Sep 7, 2009 #2
    I'd say that since the radicals are never negative, x can't be negative so you ignore the negative part and you can drop the absolute values. Don't know about the last thing.
  4. Sep 7, 2009 #3


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    Science Advisor

    In the last line, he is specifically looking to construct an open interval of x that contains x=2. Since 2 is positive, he drops the absolute value signs. By inspection, the stated internal contains x=2.
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