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Werid limit problem

  1. Jan 12, 2007 #1
    How close to 1 does x have to be to ensure that the function (x^3-1)/(x-1)^1/2 is within a distance 0.7 of its limit?

    do i find the limit as x goes to 1 then subrtace .7 or what?
     
  2. jcsd
  3. Jan 12, 2007 #2

    arildno

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    No, you are required to find a "delta" so that whenever x is closer to 1 than a distance "delta", the function value is less than "epsilon=0.7" apart from the limit value of the function at x=1.
     
  4. Jan 12, 2007 #3

    arildno

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    To help you along a bit:
    First, get rid of the denominator in the function as follows:
    [tex]f(x)=\frac{x^{3}-1}{\sqrt{x-1}}=\frac{(x^{2}+x+1)(x-1)}{\sqrt{x-1}}=(x^{2}+x+1)\sqrt{x-1}, x>1[/tex]

    Secondly, rewrite [itex]x=1+\delta, \delta>0[/itex], which is a permissible operation for any valid choice of x (there will always exist some [itex]\delta[/itex] for which that is true.

    Thirdly, rewrite the expression of f(x) in terms of [itex]\delta[/itex]:
    [tex]f(x)=f(1+\delta)=((1+\delta)^{2}+(1+\delta)+1)\sqrt{(1+\delta)-1}=(\delta^{2}+3\delta+3)\sqrt{\delta}[/tex]

    Are you following thus far?
     
  5. Jan 12, 2007 #4
    i understand what you were doing before....but what is delta exactly?
     
  6. Jan 12, 2007 #5

    HallsofIvy

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    Since [itex]x= 1+\delta[/itex], [itex]\delta= x- 1[/itex], the distance between x and 1- the number you are looking for.
     
  7. Jan 13, 2007 #6

    Gib Z

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    Can't we just find the limit, +- 0.7, then solve back for x?
     
  8. Jan 13, 2007 #7

    arildno

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    Rasine:
    I'll do this one for you, so that you can see how we do these problems.
    Now, one of the reasons why I substituted [itex]1+\delta[/itex] at the variable's place, is that we have the trivial relations:

    Take any x so that [itex]1\leq{x}<{1}+\delta[/itex], and define [itex]\delta_{x}=x-1[/itex]:
    [tex]\delta_{x}<\delta, \delta_{x}^{2}<\delta^{2},\sqrt{\delta_{x}}<\sqrt{\delta}[/tex]
    Therefore, [itex]0\leq{f(x)}<f(1+\delta)[/itex]
    Furthermore, with 0 being the limit of f at x=1, we have that:
    [tex]|f(x)-0|<|f(1+\delta)-0|=(\delta^{2}+3\delta+3)\sqrt{\delta}[/tex]
    Thus, if we can assign a value of delta so that [itex](\delta^{2}+3\delta+3)\sqrt{\delta}<0.7[/itex], then that inequality holds for any choice of x lying between 1 and [itex]1+\delta[/itex], and our proof is finished.

    Now, how do we find such a workable delta value.
    There many ways of doing this, here's perhaps the simplest one:
    If we ASSUME that [itex]\delta\leq{1}[/itex], then we have:
    [tex]\delta^{2}+3\delta+3<1^{2}+3*1+3=7[/tex]
    Hence, we have:
    [tex](\delta^{2}+3\delta+3)\sqrt{\delta}<7\sqrt{\delta}, \delta<1[/tex]
    Now, can we make [itex]7\sqrt{\delta}\leq{0.7}[/itex]?
    Indeed we can, if we set [itex]\delta\leq{0.01}[/itex]

    But, therefore, since 0.01<1, it follows that by choosing [itex]\delta=0.01[/itex], we have the inequality sequence, for every x [itex]1<{x}<{1+\delta},\delta=0.01[/itex]:
    [tex]|f(x)-0|<(\delta^{2}+3\delta+3)\sqrt{\delta}<7\sqrt{\delta}<7*0.1=0.7[/tex]
    which is our desired result.
     
    Last edited: Jan 13, 2007
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