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Use limit definition to prove

  1. Sep 13, 2009 #1
    Ues the limit definition to prove that the stated limit is correct. [tex] \lim_{x->-2} \frac{1}{x+1}=-1[/tex]. The limit def' is |f(x)-L|<epsilon if 0<|x-a|< delta. So we have [tex]|\frac{1}{x+1} + 1| < \epsilon if 0 < |x- (-2)| < \delta \mbox{ therefore } |\frac{1}{x+1}||x+2| < \epslion \mbox{ if } 0 < |x + 2| < \delta[/tex]. To bound [tex] \frac{1}{x+1} \mbox { let } -1 < \delta < 1 [/tex].[tex] -1 < x+2< 1 \mbox{ that implies } -2 < x+1 < 0 \mbox{ therefore } \frac{-1}{2} > \frac{1}{x+1} > 0[/tex]. This cannot be correct as the answer for delta is [tex] \frac{\epsilon}{1+\epsilon}[/tex]. My algebra is rusty. Please help, thanks.
     
  2. jcsd
  3. Sep 13, 2009 #2
    Not sure why you abandoned your initial line of thought, since there is no single correct answer for delta. The only problem with your analysis is that you have -2 < x + 1 < 0 and then took reciprocals, but you have 0 on one side so following the normal operations of taking reciprocals of inequalities would yield 1/(x+1) > 1/0, which is absurd. Instead, suppose you had picked delta [itex]\leq[/itex] 1/2. Then all that would change is that you now have -1/2 < x + 2 < 1/2 so -3/2 < x + 1 < -1/2, which means that 1/2 < |x+1| < 3/2 and finally 2/3 < 1/|x+1| < 2. Now you have an upper bound on 1/|x+1| without having to worry about division by zero.
     
  4. Sep 13, 2009 #3
    Using snipez90's suggestion for delta, I get [tex] \delta=min(\frac{1}{2},\frac{\epsilon}{2}) [/tex]. I was wondering how the book got the result I said it got, that is why I abandoned my original line of thought.
     
  5. Sep 14, 2009 #4

    arildno

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    John:

    When it comes to estimates and inequalities like this, you are NOT to reproduce the "answer" the book gives you (I find it scandalous and deeply anti-pedagogic that a textbook actually provides "answers" to such exercises!)

    Depending on how you think, numerous CORRECT deductions might yield very different delta-values, as functions of the epsilon.

    It can be that we may find the "optimal" (i.e, largest) delta that CAN work, but this does not in any way change the validity of a deduction leading to a smaller delta.

    Thus, there is no single, correct answer to these types of questions (there are many correct answers)
     
  6. Sep 14, 2009 #5
    I am not very familiar with the delta epsilon proof , but is it not sufficient to show that for a given delta you can find an epsilon , do we need to show the opposite ?
     
  7. Sep 14, 2009 #6
    I am sorry if I broke one of your rules. It was not my intension to do so. As I am teaching myself from a University maths book, which are generally not designed with the self taught in mind. I find that the answer to odd numbered questions is in general the only guide as to how I might be progressing. I admit that that is not ideal and that having some answers to questions can throw you (as it did here) off a valid line of reasoning or investigation. All I can do is offer my apology again.

    I am just learning the epsilon delta proof myself. Thanks for the replies.
     
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