Epsilon/delta?

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

2 Answers
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.

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