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Thread: Delta Epsilon proof

  1. #1
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    Delta Epsilon proof

    So I understand the general idea behind a delta epsilon proof but for some reason this one has me stumped. Any help would be appreciated! Thanks!

    Attached I included the question and what I have so far. I am actually stumped on how to continue from there. Thank you!
    Attached Thumbnails Attached Thumbnails Delta Epsilon proof-img_0336.jpg   Delta Epsilon proof-img_0337.jpg  
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  2. #2
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    Re: Delta Epsilon proof

    If |x+ 3|< 1 then -1< x+ 3< 1. So 3< x+ 7< 5 and \sqrt{3}< \sqrt{x+ 7}< \sqrt{5}. 2+ \sqrt{3}< \sqrt{x+ 7}+ 2< 2+ \sqrt{5}. Then \frac{1}{2+ \sqrt{5}}< \frac{1}{\sqrt{x+ 7}+ 2}< \frac{1}{2+ \sqrt{3}}.

    Replace the \frac{1}{\sqrt{x+ 7}+ 2} with othe upperbound \frac{1}{2+ \sqrt{3}}.
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  3. #3
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    Re: Delta Epsilon proof

    let $\epsilon > 0$

    $|\sqrt{x+7}-2| < \epsilon$

    $-\epsilon < \sqrt{x+7}-2 < \epsilon$

    $2- \epsilon < \sqrt{x+7} < 2+\epsilon$

    $4 - 4\epsilon + \epsilon^2 < x+7 < 4 + 4\epsilon + \epsilon^2$

    $\epsilon^2 - 4\epsilon < x+3 < \epsilon^2 + 4\epsilon$

    let $\delta = 3\epsilon$

    $|x+3| < \delta = 3\epsilon$

    $-3\epsilon < x+3 < 3 \epsilon \overset{\text{small }\epsilon}{\Rightarrow} \epsilon^2 - 4\epsilon < x+3 < \epsilon^2 + 4\epsilon \Rightarrow |\sqrt{x+7}-2| < \epsilon$

    so

    $\forall \epsilon > 0,~\exists \delta = 3\epsilon > 0 \ni |x+3|<\delta \Rightarrow |\sqrt{x+7}-2| < \epsilon$

    This is the epsilon-delta limit definition of $\displaystyle \lim_{x\to -3} \sqrt{x+7} = 2$
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