limit x - 5 (x^2 + 4) the answer listed on here from my book (c.pdf

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limit x -> 5 (x^2 + 4) the answer listed on here from my book (calc larson 9th ed) says since the interval (4,6) we have |x+5|<11 so we choose (delta) = .01/11 why is this interval chosen and how and why is |x+5|<11 Solution limit x -> 5 (x^2 + 4) here limit x ------5 means x is in between 4 and 6 4 a f(X) =L if and only if given >0 there exists >0 such that if 0<|x-a|< then |f(x)-L|< 0<|x-5|< ==> |x2+4-29|< 0<|x-5|< ==> |x2-25|< 0<|x-5|< ==> |(x-5)(X+5)|< ---------eq.2 let i took =0.01 then from eq.1 and eq.2 =0.01/11.

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