Given:
P(x) = 4+x-3x^3
We know that:
lim_(x->1) P(x) = P(1) = 2
as P(x) is a polynomial and all polynomial are continuous functions.
To prove it using the delta/epsilon method, put x=1+t, s that:
P(t) = 4+(1+t)-3(1+t)^3
P(t) = 5+t-3(1+3t+3t^2+t^3)
P(t) = 2-8t-9t^2-3t^3
Evaluate the difference:
abs (P(t) -2 ) = abs (2-8t-9t^2-3t^3-2) = abs(8t+9t^2+3t^3)
Now, for any epsilon > 0 choose delta_epsilon < min (1, epsilon/20), then we have that:
x in (1-delta_epsilon, 1+delta_epsilon) => abs(t) < delta_epsilon
Based on the triangular inequality:
abs (P(t) -2 ) <= abs(8t) + abs(9t^2)+ abs(3t^3)
and since delta_epsilon < 1 we have that abs(t) < 1 so that abs(t^3) < t^2 < abs (t)
abs (P(t) -2 ) < 8abs(t) +9 abs(t)+ 3abs(t) = 20abs(t)
abs (P(t) -2 ) < 20delta_epsilon < 20*epsilon/20 = epsilon
In conclusion we proved that if we choose delta_epsilon < min (1, epsilon/20):
x in (1-delta_epsilon, 1+delta_epsilon) => abs (P(x)-2) < epsilon
