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#
