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Epsilon delta continuity of 1/x at x=1

  1. Feb 28, 2013 #1
    This isn't really homework; It's just something that has been bothering me ever since I first learned calculus because I suck at epsilon-delta proofs.
    1. The problem statement, all variables and given/known data
    Show that 1/x is continuous at x=1


    2. Relevant equations
    If |x-a|<δ
    Then |f(x)-f(a)|<ε


    3. The attempt at a solution
    x<δ+1
    1/x<ε+1
    (ε+1)x>1
    1/(ε+1)<x<δ+1
    *starts crying*
     
  2. jcsd
  3. Feb 28, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You appear to have a confused idea of what limits are. You cannot just choose δ and
    ε independently. The definition of "limit" is
    [itex]\lim_{x\to a} f(x)= L[/itex] if and only if given [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|f(x)- L|< \epsilon[/itex].

    That is, [itex]\delta[/itex] will typically depend upon [itex]\epsilon[/itex]. The simplest way to prove a limit is to start from [itex]|f(x)- L|<\epsilon[/itex] and use that to derive a value for [itex]\delta[/itex]. As long as every step is "reversible that shows we can from [itex]|x- a|< \delta[/itex] to [itex]|f(x)- L|< \epsilon[/itex].

    Here, f(x)= 1/x, a= 1, and L= 1. So, given [itex]\epsilon> 0[/itex] we want to get [itex]|1/x- 1|< \epsilon[/itex] (notice the absolute value- you didn't have that). That is the same as [itex]-\epsilon< (1/x)- 1< \epsilon[/itex] which is the same as [itex]-\epsilon< (1- x)/x< \epsilon[/itex]. As long as x is close to 1 ([itex]\delta[/itex]< 1) x> 0 so [itex]-x\epsilon< 1- x< x\epsilon[/itex].
     
  4. Feb 28, 2013 #3

    pasmith

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    Homework Helper

    You need to show that for all [itex]\epsilon > 0[/itex] there exists [itex]\delta > 0[/itex] such that if [itex]|x - 1| < \delta[/itex] then [itex]|1/x - 1| < \epsilon[/itex].

    We will clearly need [itex]0 < \delta < 1[/itex] (or else 0 will be in our interval). Now 1/x is strictly decreasing for [itex]x > 0[/itex] (this is a consequence of [itex]\mathbb{R}[/itex] being an ordered field), so for [itex]0 < 1 - \delta < x < 1 + \delta[/itex] the following inequality is true:
    [tex]\frac{1}{1 + \delta} < \frac1x < \frac{1}{1 - \delta}[/tex]

    So if given [itex]\epsilon > 0[/itex] we can find [itex]\delta[/itex] such that
    [tex]
    1 - \epsilon < \frac{1}{1 + \delta} < \frac1x < \frac{1}{1 - \delta} < 1 + \epsilon
    [/tex]
    then we will be done. Note that if [itex]\epsilon \geq 1[/itex] then the left-most inequality is satisfied for any [itex]\delta > 0[/itex].
     
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