Epsilon-delta proof of linear eq. with negative slope

In summary, you are trying to find a \delta such that|-3| \cdot |x - 2| < \epsilon provided that 0 < |x - 2| < \delta. You already found that |-3| = 3, and |x - 2| < \epsilon / 3, so \delta = \frac{\epsilon}{3}.
  • #1
lordofpi
17
0
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?
 
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  • #2
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
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
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
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.
 

1. What is an Epsilon-delta proof?

An Epsilon-delta proof is a method used in mathematics to prove the convergence or continuity of a function. It involves choosing a value, epsilon, and showing that for any value of delta, the function will be within epsilon of the desired value.

2. How does an Epsilon-delta proof work for linear equations with negative slope?

For a linear equation with a negative slope, the epsilon-delta proof involves choosing a value, epsilon, and showing that for any value of delta, the function will be within epsilon of the desired value. This is done by finding a suitable value, delta, that will make the function approach the desired value from both sides.

3. What is the importance of using an Epsilon-delta proof for linear equations with negative slope?

An Epsilon-delta proof is important because it provides a rigorous and precise method for proving the convergence or continuity of a function. This is especially useful for linear equations with negative slope, as it allows us to determine the behavior of the function near the desired value.

4. What are the key steps in an Epsilon-delta proof of a linear equation with negative slope?

The key steps in an Epsilon-delta proof of a linear equation with negative slope include choosing a value, epsilon, and finding a suitable value, delta, that will make the function approach the desired value from both sides. Then, using algebraic manipulations, we can show that the difference between the function and the desired value is less than epsilon for any value of delta.

5. Are there any limitations to using an Epsilon-delta proof for linear equations with negative slope?

While an Epsilon-delta proof is a powerful tool, it may not always be applicable or easy to use in certain situations. Additionally, it may require advanced mathematical knowledge and skills to construct a rigorous proof. Therefore, it is important to consider other methods of proof as well.

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