Epsilon/delta?

Find and prove using the epsilon/delta definition:
#\lim _{x\to 1}\(3x^2-4)#

2 Answers
Andrea S.
Jun 15, 2017

#f(x) =3x^2-4# is a polynomial function, and as such it is defined and continuous in all of #RR#, hence:

#lim_(x->1) (3x^2-4) = f(1) = -1#

To prove it we choose arbitrarily #epsilon > 0#, and we have to prove that there exists a corresponding #delta_epsilon >0# such that for #abs(x-1) < delta_epsilon# we have:

#abs (3x^2-4+1) = abs(3x^2-3)< epsilon#

Note now that:

#abs (3x^2-3) = 3 abs (x^2-1)#

#abs (3x^2-3) = 3abs((x-1)(x+1))#

#abs (3x^2-3) = 3abs(x-1)abs(x+1)#

and, if #delta_epsilon < 1#, for #x in (1-delta_epsilon,1+delta_epsilon)# clearly:

#abs (x-1) < delta_epsilon#

#abs(x+1) < 2+delta_epsilon < 3#

so,

#abs (3x^2-3) < 3 delta_epsilon (2+delta_epsilon) < 9delta_epsilon#

Then if for any #epsilon >0# we choose #delta_epsilon < min(1, epsilon/9)# we have that:

#abs (3x^2-3) < epsilon#

which proves the point.

Jim H
Jun 17, 2017

Please see below.

Explanation:

Claim: #lim_(xrarr1)(3x^2-4) = -1#

For #epsilon > 0#, let #delta = min{1, epsilon/9}#.

Observe that if #0 < abs(x-1) < delta#, then #

#abs(x-1) < 1#, so that #-1 < x-1 < 1# and thus #1 < x+1 < 3# allowing us to conclude that #abs(x+1) < 3#

Furthermore, (for #abs(x-1) < delta#), we have

#abs((3x^2-4)-(-1)) = abs(3x^2-3)#

# = 3abs(x^2-1)#

# = 3abs(x-1)abs(x+1)#

# < 3(delta)(3)#

# = 9(epsilon/9)#

# = epsilon#

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