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Homework Help: Limit prooving question

  1. Jan 6, 2009 #1
    the question and where i got stuck in this link:

    http://img187.imageshack.us/img187/8868/44429492ix7.gif [Broken]
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 6, 2009 #2


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    You have a proof that limit, as x goes to 5 of x2= 25. For that you must have [itex]|x^2- 25|< \epsilon[/itex] which leads to [itex]|(x- 5)(x+ 5)|< \epsilon[/itex]
    Since you want something of the form [itex]|x- 5|< \delta[/itex], is is the "x+ 5" term you want to deal with. You cannot write [itex]|x- 5|< \epsilon/(x+5)[/itex] because the right side cannot depend on x. What you can do is find an upper bound: if |x+5|< M, then 1/|x+5|> 1/M so [itex]\epsilon/|x+5|> \epsilon/M[/itex]. If, then [itex]|x-5|< \epsilon/M[/itex] which is itself less than [itex]\epsilon/|x+5|[/itex] it is certainly less than [itex]\epsilon/(x+5)[/itex]

    What is an upper bound on |x+5|? First you must understand that there are an infinite number of such "upper bounds". If a is an upper bound then any number larger than a is still an upperbound. We want x close to 5. Just because "1" is an easy number, lets try |x-5|< 1. That says that -1< x- 5< 1 and, adding 10 to each part 9< x+5< 11. Since 9< x+ 5, it is obviously positive so |x+5|= x+ 5< 11. I chose "10" to add, of course, because x-5+ 10= x+ 5.

    That's where those numbers come from. We could as easily have decided that |x-5|< 2 is "close enough". Then we would have -2< x- 5< 2 and so, adding 10 again, 8< x+ 5< 12 and now we "12" as an upper bound on |x+5| rather than 11.

    Or we could decide we wanted to be |x-5|< 1/2. -1/2< x- 5< 1/2. Again adding 10, 9.5< x+ 5< 10.5 so that 10.5 is an upper bound. In fact, any number greater than 10 could be used as an upper bound.
  4. Jan 6, 2009 #3
    Thanks i understand that
  5. Jan 6, 2009 #4
    i understand that our upper bound is 11

    so inorder for this to work we need only

    [itex]\delta = \frac{\epsilon }{11}

    why to use minimum??

    why they are taking the solution as the minimal between 1 and the other value??

    [itex]\delta =\min \left\{ 1,\frac{\epsilon }{11} \right\}.\quad\blacksquare
  6. Jan 7, 2009 #5


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    We have shown that in order that |x- L|< [itex]\epsilon[/itex], we must have |x- a|, [itex]\epsilon/11[/itex]. But in order to have that we must have |x-a|<1. That is, they must both be true.

    If |x-a| is less than the smaller of the two, it is less than both.
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