Epsilon-delta proof of linear eq. with negative slope

[lordofpi]

I am familiar with most of how to do ε-δ proofs (even though our professor thought it unimportant to teach it, and our book kind of glosses over it (Larson, Fundamentals of Calculus, 9th), even quadratically, but for some reason I am just getting stuck on what is probably a simple problem.

Homework Statement

Find L. Then find ε > 0 and δ > 0 to satisfy the definition of a limit.
Given: \lim_{x \to 2} 5 - 3x

2. The attempt at a solution

First I determine \lvert x-2 \rvert < \delta.

Then, I solved for the limit L analytically
\lim_{x \to 2} 5 - 3x = 5 - 3(2) = -1

Now given that f(x) - L < \epsilon,
this limit can ben stated as \lvert 5 -3x -(-1) \rvert < \epsilon
Which can be restated as \lvert -3(x-2)\rvert < \epsilon
Or even \lvert -3 \rvert \cdot \lvert x-2 \rvert < \epsilon

So now what do I do? I am aware that \lvert -3 \rvert = 3, but to jump to that just to get an easy \frac{\epsilon}{3} value for \delta just seems improper. Plus I don't know how one can properly reintroduce the minus sign back into the absolute value when doing the final proof (not shown). Any thoughts on what I am overlooking?

 

[jbunniii]

Your work so far seems fine. Your goal is to find a \delta > 0 such that
|-3| \cdot |x - 2| < \epsilon
provided that 0 < |x - 2| < \delta.

You already noted that |-3| = 3, so the inequality reduces to
3|x-2| < \epsilon
which is of course equivalent to
|x - 2| < \epsilon / 3
Now what value of \delta can you choose to guarantee that this inequality will hold, as long as 0 < |x - 2| < \delta?

 

[lordofpi]

Okay, so provided that is all legal, then:

If \lvert x-2 \rvert < \delta
And \delta = \frac{\epsilon}{3}

Then \lvert x-2 \rvert &lt; \frac{\epsilon}{3}\\<br /> \downarrow\\<br /> 3 \cdot \lvert x-2 \rvert &lt; \epsilon\\<br /> \downarrow\\<br /> \lvert 3x - 6\rvert &lt; \epsilon\\<br /> \downarrow\\<br /> \lvert3x - 5 - 1\rvert &lt; \epsilon\\<br /> \downarrow

And now for the step I was afraid to do:
\lvert-3x + 5 + 1 \rvert&lt; \epsilon\\<br /> \downarrow\\<br /> \lvert 5-3x-(-1) \rvert&lt; \epsilon\\<br /> \downarrow\\<br /> \lvert f(x) - L\rvert &lt; \epsilon
Q.E.D.

Is that correct?

 

[jbunniii]

Yes, that's entirely correct. There's nothing wrong with the step that you were afraid to do. |-(anything)| = |anything|, regardless of what "anything" is.

 

[lordofpi]

Great. Thanks so much for putting my mind at ease. It's your statement in your last post that I just wasn't sure of. While it was intuitively correct, math doesn't always work that way, and I didn't want to take any chances of teaching myself some fouled up shortcut :). Funny enough, I just came across a sample \epsilon-\delta problem in my chapter review of \lim_{x \to 2}1-x^2 and cut through it like butter.

Also, I wasn't sure I could just manipulate the proof any way I chose -- but I guess that is sort of the nature of a proof, as long as the manipulations are mathematically sound. Thanks again.