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.

1. 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