Evaluate the difference:
#abs(5x^2-5) = 5abs(x^2-1)#
#abs(5x^2-5) = 5abs(x-1)abs(x+1)#
Now for any #epsilon > 0# choose #delta_epsilon < min(1,epsilon/15)#.
As #delta_epsilon < 1# then for #x in (1-delta_epsilon, 1+delta_epsilon)# we have that:
#0 < x < 2#
and:
#1 < 1+x < 3#
Then #abs(1+x) = (1+x) < 3# and so:
#abs(5x^2-5) = 5abs(x-1)abs(x+1) < 15abs(x-1)#
On the other hand, as #delta_epsilon < epsilon/15# then for #x in (1-delta_epsilon, 1+delta_epsilon)# we have that:
#abs(x-1) < delta_epsilon < epsilon/15#
So:
#abs(5x^2-5) < 15abs(x-1)#
#abs(5x^2-5) < 15epsilon/15#
#abs(5x^2-5) < epsilon#
which proves the limit.
