# Epsilon-delta proof of linear eq. with negative slope

1. Sep 21, 2012

### 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.

1. The problem statement, all variables and given/known data

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?

2. Sep 21, 2012

### 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 $0 < |x - 2| < \delta$?

Last edited by a moderator: Sep 21, 2012
3. Sep 21, 2012

### lordofpi

Okay, so provided that is all legal, then:

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

Then $$\lvert x-2 \rvert < \frac{\epsilon}{3}\\ \downarrow\\ 3 \cdot \lvert x-2 \rvert < \epsilon\\ \downarrow\\ \lvert 3x - 6\rvert < \epsilon\\ \downarrow\\ \lvert3x - 5 - 1\rvert < \epsilon\\ \downarrow$$

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

Is that correct?

4. Sep 21, 2012

### 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.

5. Sep 22, 2012

### 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.