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Thread: 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:
    $\displaystyle \forall{\epsilon>0}\wedge\forall{x}\quad 0<|x-2|<\epsilon^2\implies |g(x)-4|<\epsilon$

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

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

    Here I suppose I need to bound $\displaystyle g(x)$ 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 $\displaystyle \varepsilon<1$. If $\displaystyle 0<|x-2|<\varepsilon^2$ then $\displaystyle 3\leq g(x)\leq 5$.
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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 $\displaystyle \varepsilon<1$. If $\displaystyle 0<|x-2|<\varepsilon^2$ then $\displaystyle 3\leq g(x)\leq 5$.
    Why? Don't I have to do it for all $\displaystyle \epsilon>0$?
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    Re: Limits (Spivak's problem)

    If you can show the result for $\displaystyle \varepsilon \leq 1$ and you take $\displaystyle \varepsilon>1$, then the $\displaystyle \delta$ which works for $\displaystyle \varepsilon=1$ will work for this $\displaystyle \varepsilon$.
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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 $\displaystyle \epsilon>0$?
    Start off with any $\displaystyle \varepsilon > 0$ let $\displaystyle \varepsilon ' = \min \left\{ {1,\varepsilon } \right\}$

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

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

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