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Math Help - I will use a 'g' for delta and a small 'e' for episilon to find delta

  1. #1
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    I will use a 'g' for delta and a small 'e' for episilon to find delta

    I must find the value of 'g'>0 such that |f(x) - L| < 'e' when 0 < |x-a| < 'g' lim x-> 9 (square root 3) = 3, 'e'= 0.01. I think the function of |x^(1/2 -3|< 0.01 is the proper way to set up this problem. I am having trouble at this point Generally for all of the problems I have done so far included |x-9| as x -> 9. Then I find -1 < |x-9| < 1 -> which results in 8< x< 10. Then I substitue whatever the adsolute value of (squrae root of 3 < 10. This is where I am stuck. I do not understand how to maniputae (square root 3) to substitue it in for the x in 8 < |(square root 3)|< 10. I think if I get assistance finding out the substitute absolute value I may be able to solve, and find 'g' I have tries to square square (square root x -3) to get 8 <|x-3|< 10then use the 'g' = 0.01/ 13; but that doesn't match with the answer in the back of my text. The answer in the back of my book is 0.058. I am getting 0.00076. Can someone please give me some guidance?
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  2. #2
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    First surely you mean \lim _{x \to 9} \sqrt x  = 3?
    Take note that \left| {\sqrt x  - 3} \right| = \frac{{\left| {x - 9} \right|}}<br />
{{\left| {\sqrt x  + 3} \right|}}.
    So \left| {x - 9} \right| < 1\, \Rightarrow \,\sqrt 8  < \sqrt x  < 10\, \Rightarrow \,\frac{1}<br />
{{\sqrt x  + 3}} < \frac{1}{{\sqrt 8  + 3}}
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  3. #3
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    Quote Originally Posted by Plato View Post
    First surely you mean \lim _{x \to 9} \sqrt x  = 3?
    Take note that \left| {\sqrt x  - 3} \right| = \frac{{\left| {x - 9} \right|}}<br />
{{\left| {\sqrt x  + 3} \right|}}.
    So \left| {x - 9} \right| < 1\, \Rightarrow \,\sqrt 8  < \sqrt x  < 10\, \Rightarrow \,\frac{1}<br />
{{\sqrt x  + 3}} < \frac{1}{{\sqrt 8  + 3}}
    Plato, Thank you very much for your instruction. This is great, this is great.
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