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Math Help - Limits (Spivak's problem)

  1. #1
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    Limits (Spivak's problem)

    Hi everybody, could anyone help me with this?:
    Suppose that the function g has the following property:
    \forall{\epsilon>0}\wedge\forall{x}\quad 0<|x-2|<\epsilon^2\implies |g(x)-4|<\epsilon

    For each \epsilon>0 find a \delta>0 such that, for all x:
    If 0<|x-2|<\delta, then \left|\dfrac{1}{g(x)}-\dfrac{1}{4}\right|<\epsilon

    Well, I've started working here to find an expression for \delta in terms of \epsilon:
    \left|\dfrac{1}{g(x)}-\dfrac{1}{4}\right|=\left|\dfrac{4-g(x)}{4g(x)}\right|=\dfrac{|4-g(x)|}{4|g(x)|}=\dfrac{|g(x)-4|}{4|g(x)|}

    Here I suppose I need to bound g(x) but, how? If not, what's next?
    Thanks in advance.
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    Super Member girdav's Avatar
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    Re: Limits (Spivak's problem)

    You only have to show the result for \varepsilon<1. If 0<|x-2|<\varepsilon^2 then 3\leq g(x)\leq 5.
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    Re: Limits (Spivak's problem)

    Hi, thanks for replying.
    Quote Originally Posted by girdav View Post
    You only have to show the result for \varepsilon<1. If 0<|x-2|<\varepsilon^2 then 3\leq g(x)\leq 5.
    Why? Don't I have to do it for all \epsilon>0?
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    Super Member girdav's Avatar
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    Re: Limits (Spivak's problem)

    If you can show the result for \varepsilon \leq 1 and you take \varepsilon>1, then the \delta which works for \varepsilon=1 will work for this \varepsilon.
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    Re: Limits (Spivak's problem)

    Quote Originally Posted by MATHNEM View Post
    Why? Don't I have to do it for all \epsilon>0?
    Start off with any  \varepsilon  > 0 let  \varepsilon ' = \min \left\{ {1,\varepsilon } \right\}

    Now we have 3<g(x) or \frac{1}{4g(x)}< \frac{1}{12} .

    Now you can pick \delta  = 144(\varepsilon ')^2

    So because  \varepsilon' \le \varepsilon you have done it for any  \varepsilon .
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