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Epsilon-delta proof of linear eq. with negative slope

  • Thread starter lordofpi
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  • #1
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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 [itex]L[/itex]. Then find ε > 0 and δ > 0 to satisfy the definition of a limit.
Given: [tex]\lim_{x \to 2} 5 - 3x [/tex]

2. The attempt at a solution

First I determine [itex]\lvert x-2 \rvert < \delta[/itex].

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

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

So now what do I do? I am aware that [itex]\lvert -3 \rvert = 3[/itex], but to jump to that just to get an easy [itex] \frac{\epsilon}{3}[/itex] value for [itex]\delta[/itex] 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?
 

Answers and Replies

  • #2
jbunniii
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Your work so far seems fine. Your goal is to find a [itex]\delta > 0[/itex] such that
[tex]|-3| \cdot |x - 2| < \epsilon[/tex]
provided that [itex]0 < |x - 2| < \delta[/itex].

You already noted that [itex]|-3| = 3[/itex], so the inequality reduces to
[tex]3|x-2| < \epsilon[/tex]
which is of course equivalent to
[tex]|x - 2| < \epsilon / 3[/tex]
Now what value of [itex]\delta[/itex] can you choose to guarantee that this inequality will hold, as long as [itex]0 < |x - 2| < \delta[/itex]?
 
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  • #3
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Okay, so provided that is all legal, then:

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

Then [tex]\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[/tex]

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

Is that correct?
 
  • #4
jbunniii
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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
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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 [itex]\epsilon[/itex]-[itex]\delta[/itex] problem in my chapter review of [itex]\lim_{x \to 2}1-x^2[/itex] 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.
 

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