How do you prove the statement lim as x approaches 4 for #(7 – 3x) = -5# using the epsilon and delta definition?

Our preliminary work

We need to show that, given any positive #edpsilon#, there is a #delta# (also positive) that guarantees that if #x# is chosen so that

#0 < abs(x-4) < delta#, then we will also have #abs((7-3x)-(-5))< epsilon#.

Let's look at the thing we need to make less than #epsilon#.

#abs((7-3x)-(-5)) = abs(7-3x+5)#

# = abs(12-3x)#

Because we say how to find #delta#, that means we control the sive of #abs(x-4)#

Not that if we factor out a #-#, we can get

#abs((7-3x)-(-5)) = abs(12-3x) = abs(-3(x-4))#

# = abs(-3)abs(x-4) = 3abs(x-4)#

We have found that #abs((7-3x)-(-5))# which we want to make less that #epsilon#, is equal to #3abs(x-4)#, which is #3# times the thing we control.

If we make #abx(x-4) < epsilon/4#, that will work.

Now we need to present ourwokr as a proof.

Proof

Given #epsilon > 0#, choose #delta = epsilon/4#. Now if #x# is chosen so that #0 < abs(x-4) < delta#, then we have

#abs((7-3x)-(-5)) = abs(12-3x) = abs(-3)abs(x-4) #

# = 3abs(x-4) < 3 delta = 3(epsilon/3) = epsilon.#.

That is, #abs((7-3x)-(-5)) < epsilon.#.

Note that
near the middle we used, without mention, the fact that

#abs(x-4) < delta# implies that #3abs(x-4) < 3delta#.