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Thread: Epsilon Delta proof of sqrt limit

  1. #1
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    Epsilon Delta proof of sqrt limit

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

    Proof so far.

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

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

    Now, I know that $\displaystyle \mid \sqrt {x} -2 \mid \mid \sqrt {x} + 2 \mid = \mid x -4 \mid $, so that $\displaystyle \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 $\displaystyle \lim _{x \rightarrow 4} \sqrt {x} = 2 $

    Proof so far.

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

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

    Now, I know that $\displaystyle \mid \sqrt {x} -2 \mid \mid \sqrt {x} + 2 \mid = \mid x -4 \mid $, so that $\displaystyle \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 $\displaystyle \frac{|x-4|}{\sqrt{x}+4}\leqslant\frac{|x-4|}{\sqrt{3}+4}$
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  3. #3
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    $\displaystyle \lim_{x\to 4}\sqrt{x}=2$

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

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

    $\displaystyle |x-4|<4$

    $\displaystyle 0<x<8$

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

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

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

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

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

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

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

    See here for a very nice paper on epsilon delta limits

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