How do you use the epsilon delta definition to prove that the limit of 2x-4=6 as x->1?

1 Answer
May 9, 2017

It cannot be proven. It is false.

Explanation:

lim_(xrarr1)(2x-4) = -2 and -2 != 6

Proof that lim_(xrarr1)(2x-4) = -2

Given epsilon > 0, choose delta = epsilon/2 (Note that delta > 0.)

Then for any x with 0 < abs (x-1) < delta,

abs((2x-4)-(-2)) = abs(2x-2)

= 2abs(x-1)

< 2delta

= 2epsilon/2

= epsilon

That is: abs(2x-4)-(-2)) < epsilon.

We have shown that:

For any epsilon > 0, there is a delta > 0 such that

for all x with 0 < abs(x-1) < delta we have abs((2x-4)-(-2)) < epsilon.

So, by the definition of limit, lim_(xrarr3)(2x-4) = -2.