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Math Help - Epsilon Delta proof of sqrt limit

  1. #1
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    Epsilon Delta proof of sqrt limit

    Prove that  \lim _{x \rightarrow 4} \sqrt {x} = 2

    Proof so far.

    Given  \epsilon > 0 , let  \delta > 0 (or some other positive values that I haven't figure out yet)

    Suppose that  \mid x-4 \mid < \delta , I need to show that  \mid \sqrt {x} - 2 \mid < \epsilon

    Now, I know that  \mid \sqrt {x} -2 \mid \mid \sqrt {x} + 2 \mid = \mid x -4 \mid , so that  \mid \sqrt {x} -2 \mid = \frac { \mid x-4 \mid }{ \mid \sqrt {x} + 2 \mid }

    But how would I go from here? thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tttcomrader View Post
    Prove that  \lim _{x \rightarrow 4} \sqrt {x} = 2

    Proof so far.

    Given  \epsilon > 0 , let  \delta > 0 (or some other positive values that I haven't figure out yet)

    Suppose that  \mid x-4 \mid < \delta , I need to show that  \mid \sqrt {x} - 2 \mid < \epsilon

    Now, I know that  \mid \sqrt {x} -2 \mid \mid \sqrt {x} + 2 \mid = \mid x -4 \mid , so that  \mid \sqrt {x} -2 \mid = \frac { \mid x-4 \mid }{ \mid \sqrt {x} + 2 \mid }

    But how would I go from here? thanks.
    Restrict your attention to the open ball of radius one around four so that \frac{|x-4|}{\sqrt{x}+4}\leqslant\frac{|x-4|}{\sqrt{3}+4}
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  3. #3
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    \lim_{x\to 4}\sqrt{x}=2

    |\sqrt{x}-2|=\left|\frac{\sqrt{x}-2}{1}\cdot \frac{\sqrt{x}+2}{\sqrt{x}+2}\right|=\frac{|x-4|}{\sqrt{x}+2}

    If we restrict {\delta} so that {\delta}\leq 4, then

    |x-4|<4

    0<x<8

    0<\sqrt{x}<\sqrt{8}

    2<\sqrt{x}+2<\sqrt{8}+2

    \frac{1}{2}>\frac{1}{\sqrt{x}+2}>\frac{1}{\sqrt{8}  +2}

    \frac{1}{\sqrt{x}+2}<\frac{1}{2}

    \frac{|x-4|}{\sqrt{x}+2}\leq \frac{1}{2}|x-4|

    Therefore, |\sqrt{x}-2|<{\epsilon} if \frac{1}{2}|x-4|<{\epsilon}, \;\ or \;\ |x-4|<2{\epsilon}

    So, {\delta}=min(2{\epsilon},4)

    See here for a very nice paper on epsilon delta limits

    http://www.mathhelpforum.com/math-he...ta-proofs.html
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