Verified answer

When you have to do an ε-δ (or ε-N) proof, you pretty much always start with "Let ε > 0." In fact, my Probability Professor told of how she would preprint sheets of paper with exactly that at the top for when she had to do such a proof.

After you introduce ε, you need to then introduce delta, which generally should be a function of ε. The way in which you calculate δ does not need to be shown in the proof. That is side/scratch work. All you need to do is introduce δ, and do some quick algebra.

The dirty work that you do behind the scenes (and do not put in the proof!), as it were, is take |f(x) - L| < ε, then do some algebra to figure out what δ for which |x - c| < δ, which will usually be an expression involving ε. This is basically working backwards.

Then when you go back to writing the proof, you introduce δ as such, assume that 0 < |x - c| < δ, and show that |f(x) - L| < ε, and this is the process by which you show that lim[x -> c] f(x) = L.

#1

lim of (f(x)) as x approches a is L.

the def. is: (for every ε >0)(there exist a δ >0)s.t. if 0<|x-a|<δ then |f(x)- L|<ε

applying:

(for every ε >0)(there exist a δ >0)s.t. if 0<|x-1|<δ then |(3x-7)+4|<ε

let ε >0.

Take δ = ε /3.

|(3x-7)+4| = |3x-3| = |3||x-1| < 3δ = 3(ε /3) = ε

therefore, the lim (x -> 1) (3x - 7) = -4.

*not part of the proof:

you get the δ from solving |x-1|<δ to get |(3x-7)+4|<ε

multiplying both sides of inequality by 3

we get |3x-3|<3δ which is already equal to |3x-7+4| which is <ε

so we take 3δ = ε

solving for δ we get δ = ε/3.

do the same process for #2.

Let f(x) = (3x - 7)

We want to prove that

lim (3x - 7) = -4

x -> 1

by proving that for every ε > 0, there exists δ > 0 such that whenever |x - a| < δ, |f(x) - L| < ε

i.e. we want to prove that for every ε > 0, there exists δ > 0 such that whenever |x - 1| < δ, |3x - 7 - (-4)| < ε

-OR-

|x - 1| < δ, |3x - 3| < ε

It is as simple as choosing δ = (1/3)ε, since, if

|x - 1| < (1/3)ε, then

3|x - 1| < ε

|3(x - 1)| < ε

|3x - 3| < ε

|3x - 7 = (-4)| < ε

Showing that

whenever |x - 1| < δ, |3x - 7 - (-4)| < ε