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Math Help - One more epsilon ?

  1. #1
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    Dec 2009
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    One more epsilon ?

    I am given

    f(x) = sq root x
    Xo= 1
    L=1
    E= 1/4
    y= sq root x

    and it says

    find a delta > 0 that for all x 0< abs value x-Xo < alpha -> abs value f(x)-L < E

    I know on the right side I will have sq root x - 1 < 1/4

    I am not sure what I put on the left side however. I know what Xo is but what about the x and the alpha?

    I am so confused
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  2. #2
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    Quote Originally Posted by KarlosK View Post
    I am given

    f(x) = sq root x
    Xo= 1
    L=1
    E= 1/4
    y= sq root x

    and it says

    find a delta > 0 that for all x 0< abs value x-Xo < alpha -> abs value f(x)-L < E

    I know on the right side I will have sq root x - 1 < 1/4

    I am not sure what I put on the left side however. I know what Xo is but what about the x and the alpha?

    I am so confused
    You want to find a δ>0 such that :

    for all ,x : if o<|x-1|< δ ,then |\sqrt{x}-1|<\frac{1}{4}.................................................. ............................................1

    But |\sqrt{x}-1| = \frac{|(\sqrt{x}-1)(\sqrt{x}+1)|}{|\sqrt{x}+1|}=\frac{|x-1|}{\sqrt{x}+1} \leq|x-1|.................................................. ................................................2

    Since \sqrt{x}\geq 0\Longrightarrow\sqrt{x}+1\geq 1\Longrightarrow\frac{1}{\sqrt{x}+1}\leq 1 \Longrightarrow\frac{|x-1|}{\sqrt{x}+1}\leq |x-1|

    So from (2) we see that if we put |x-1|<1/4 ,then |\sqrt{x}-1|<\frac{1}{4}.

    Obviously δ = 1/4
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