How do you use the epsilon delta definition to prove that the limit of #sqrt(x+6)=9# as #x->3#?

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Answer 1 · 2018-05-23T19:00:31

For each epsilon, we can always choose #delta < (9 + epsilon)^2 - 9#

#forall epsilon >0, exists delta >0 ; | x - 3| < delta => |sqrt{x + 6} - 9| < epsilon#

I will think backwards

#-epsilon < sqrt{x + 6} - 9 < epsilon#

plus nine

#9-epsilon < sqrt{x + 6} < 9 + epsilon#

^2

#(9-epsilon)^2 < x + 6 < (9 + epsilon)^2#

minus 9

#(9-epsilon)^2 - 9 < x - 3 < (9 + epsilon)^2 - 9#

Why? All we want is #-delta < x - 3 < delta#

We can always choose #delta < (9 + epsilon)^2 - 9#

And #(9 - epsilon)^2 - 9 < - delta#

#9 - (9 - epsilon)^2 < delta < (9 + epsilon)^2 - 9#

#9 - (9 - epsilon)^2 < (9 + epsilon)^2 - 9#

#18 < (9 + epsilon)^2 + (9 - epsilon)^2 = 81 + 81 +(18-18)epsilon + 2epsilon^2#

#9-81 = -72 < epsilon^2#