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Another epsilon-delta proof

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that

    [tex]
    \lim_{x \to 3} \frac{1}{x-4} = -1
    [/tex]

    2. Relevant equations



    3. The attempt at a solution

    Given [tex]\varepsilon > 0[/tex], we want to find [tex]\delta[/tex] such that

    [tex]
    \begin{align*}
    0<|x-3|<\delta \Rightarrow |\frac{1}{x-4}+1| < \varepsilon
    \end{align*}
    \begin{align*}
    |\frac{1}{x-4}+1| &< \varepsilon\\
    |\frac{x-3}{x-4}| &< \varepsilon
    \end{align*}
    [/tex]

    I always get stuck at this part! I'm never sure of how to proceed. Should I restrict the interval?
     
  2. jcsd
  3. Oct 25, 2009 #2

    Mark44

    Staff: Mentor

    You could could restrict |x - 3| to be less than .5, which means that 2.5 < x < 3.5, and -1.5 < x - 4 < -.5, so .5 < |x - 4| < 1.5.
     
  4. Oct 25, 2009 #3
    so once we get to here, using [tex]\delta_1 = \frac{1}{2}[/tex],

    [tex]
    \begin{align*}
    \frac{1}{2} < |x-4| < \frac{3}{2}\\
    2 > |\frac{1}{x-4}| > \frac{2}{3}
    \end{align*}
    [/tex]

    and then plugging it into the epsilon formula

    [tex]
    \begin{align*}
    |\frac{x-3}{x-4}| < \varepsilon\\
    2|x-3| < \varepsilon\\
    |x-3| < \frac{\varepsilon}{2}
    \end{align*}
    [/tex]

    so we take

    [tex]\delta = \text{min} (\frac{1}{2}, \frac{\varepsilon}{2})[/tex]

    does it seem okay?
     
  5. Oct 26, 2009 #4
    sorry, i never got a response on this one, and i'm still not sure if it's right.
     
  6. Oct 26, 2009 #5
    |(x-3)/(x-4)| < epsilon, |x-3| =/= 1, take delta = 1/2 --> - 1/2 < x-3 < 1/2, 5/2 < x < 3 7/2 --> -3/2 < x - 4 < -1/2 --> |x-4| > 1/2 --> |(x-3)/(x-4)| < 2|(x-3)| < epsilon


    in short, delta = min(1/2 , epsilon/2)
     
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