Verified answer

lim(x-->x₀) f(x)=A

or

∀ε>0 ∃δ>0: ∀x∈U(δ,x₀) => f(x)∈U(ε,A)

Let ε=1

[(√x)-4]∈U(1,1) <=> |(√x)-4 - 1|<1

|(√x)-4 - 1|<1

|(√x)-5|<1

-1<(√x)-5<1

4<√x<6

16<x<36

16-25<x-25<36 -25

-9<x-25<11

Let -9<x-25<9 (<11)

-9<x-25<9

|x-25|<9

Hence

δ=9

Note that

√(x) - 4 - 1 = √(x) - 5 = (x - 25)/(√(x) + 5)

If |x - 25| < δ and δ ≤ 9, then

16 < x < 34 ==> 4 + 5 < √(x) + 5 < √(34) + 5 ==> 1/(√(34) + 5) < 1/(√(x) + 5) < 1/9.

In this case,

|√(x) - 5| < δ/9

This is less than ε provided δ = min{9, 9ε}. So for ε = 0.1, take δ ≤ 0.9.