Let y = L[x] denote the tangent line to the function y = f[x] = (Sqrt[1+(x^3)]) at (1, f[1]) = (1, (Sqrt[2])).
The following finds an x--interval around x = 1 on which  | L[x] - f[x] | < .01

f[x_] = Sqrt[x^3 + 1]
ϵ = .01
fplus[x_] := f[x] + ϵ
fminus[x_] := f[x]-ϵ

Tangent line at  x = 1:

L[x_] = Sqrt[2] +  (3/(2Sqrt[2])) * (x -1)

d = .2
Plot[{fplus[x], fminus[x], L[x]}, {x, 1-d, 1 + d}]

d = .1
Plot[{fplus[x], fminus[x], L[x]}, {x, 1-d, 1 + d}]

We see from this that  the interval of radius .2 is too large to have the desired property but the interval of radius .1 works.

This was the graphic method suggested in the text, but we can see things more clearly if instead we consider
(and plot) the difference L[x] - f[x],  which is, after all the thing we want to make small

ϵ = .01;
η[x_] :=  L[x] - f[x]
d = .1;
Plot[{ϵ, -ϵ, η[x]}, {x, 1-d, 1 + d}]

ϵ = .0001;
η[x_] :=  L[x] - f[x]
d = .01;
Plot[{ϵ, -ϵ, η[x]}, {x, 1-d, 1 + d}]