Prove that limx→ 2 (3x-2)=4 using epsilon,delta function definition

Answer 1

lim f(x) = L as x->t if for every e>0 there exists h>0 such that

|f(x)-L|<e whenever |x-t|<h

working backwards

| 3x-2 -4 | < e

|3x - 6| < e

3|x-2| < e

|x-2| < e/3

-e/3 < x-2 < e/3

2 - e/3 < x < 2 + e/3

this suggests we choose h=e/3

Since e>0, then h >0/3 = 0

So when |x-2| < h=e/3

3|x-2| < e

|3||x-2| < e

|3x-6| < e

-e < 3x-6 < e

4-e < 3x-2 < 4+e

4-e < f(x) < 4+e

So when x is in a neighborhood of 2, the function

is in a neighborhood of 4

[end of proof]