Epsilon-Delta Proof (Right or Wrong)?

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In summary: DIn summary, the $\epsilon-\delta$ definition of a limit states that for any given $\epsilon>0$, there exists a corresponding $\delta>0$ such that if $|x-3|<\delta$, then $|f(x)-\frac{1}{3}|<\epsilon$. To prove this, we can fix an arbitrary $\epsilon>0$ and consider $\delta=\min(1,15\epsilon)$. Then, if $|x-3|<\delta$, we have $|f(x)-\frac{1}{3}|<\epsilon$ as desired. This method simplifies the process by finding an upper bound for $|x+3|$ and using it to find
  • #1
bwpbruce
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1
Given:
\(\displaystyle \lim_{x \to 3} \dfrac{2}{x + 3} = \dfrac{1}{3}\)

Goal:
Prove using $\epsilon-\delta$ definition of a limit.

Plan:
Let \(\displaystyle \epsilon > 0 \) be given. Find \(\displaystyle \delta > 0 \) so that
if \(\displaystyle 0 < \left|x - 3 \right| < \delta \), then \(\displaystyle \left|f(x) - \dfrac{1}{3} \right| < \epsilon \)

Solve:
\(\displaystyle \left|\dfrac{2}{x + 3} - \dfrac{1}{3} \right| = \left|\dfrac{6}{3(x + 3)} - \dfrac{x + 3}{3(x + 3)} \right| = \left| \dfrac{x + 3}{3(x + 3)} - \dfrac{6}{3(x + 3)}\right| = \left| \dfrac{x + 3 - 6}{3(x + 3)}\right|= \left| \dfrac{x -3}{3(x + 3)}\right| < \epsilon \)
\begin{align*} |x - 3| &< |3(x + 3)|\epsilon \\ x\text{ is approaching }3, \text{therefore }|x - 3| &< |3(3 + 3)|\epsilon \\ |x - 3| &< |3(6)| \\ \left|x - 3 \right| &< 18\epsilon\end{align*}

Result:
\(\displaystyle 18\epsilon = \delta \)

Conclusion:
From the statement \(\displaystyle 18\epsilon = \delta \), it is clear that for any given \(\displaystyle \epsilon\), a \(\displaystyle \delta\) exists such that if \(\displaystyle 0 < \left|x - 3 \right| < \delta \), then \(\displaystyle \left|f(x) - \dfrac{1}{3} \right| < \epsilon \)
 
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  • #2
So, does this mean that for $\epsilon=1$ I can take $\delta=18$? Let's take $x=-2.99$, then $|x-3|=5.99<\delta$, but $\left|\dfrac{2}{x+3}-\dfrac{1}{3}\right|=\left|\dfrac{2}{0.01}-\dfrac{1}{3}\right|=\left| 200-\dfrac{1}{3}\right|>\epsilon$.

I recommend writing your solution not as a record of finding $\delta$ (i.e., not as a process of solving \(\displaystyle \left|f(x) - 1/3 \right| < \epsilon \)), but as a proof that a certain $\delta$ works. To illustrate the difference, your solution contains sequences of statements without describing the relationship between them, for example:
bwpbruce said:
\begin{align*} |x - 3| &< |3(x + 3)|\epsilon \\ x\text{ is approaching }3, \text{therefore }|x - 3| &< |3(3 + 3)|\epsilon \\ |x - 3| &< |3(6)| \\ \left|x - 3 \right| &< 18\epsilon\end{align*}
Does this mean that $|x - 3| < |3(x + 3)|\epsilon$ implies $\left|x - 3 \right| < 18\epsilon$ or vice versa? Since the definition of limit says $|x-3|<\delta\implies |f(x)-1/3|<\epsilon$, it must be the case that $\left|x - 3 \right| < 18\epsilon$ implies $|x - 3| < |3(x + 3)|\epsilon$. But even if $x<3$ (and the fact that $x$ approaches 3 does not mean that $x<3$), then $|x - 3| < |3(3 + 3)|\epsilon$ does not imply $|x - 3| < |3(x + 3)|\epsilon$. The converse implication holds: $|x - 3| < |3(x + 3)|\epsilon<|3(3 + 3)|\epsilon$, but even this requires that $-3<x<3$.

A good proof should start with assumptions ("Let's fix an arbitrary $\epsilon>0$ and consider $\delta=\ldots$") and then show that they necessarily imply \(\displaystyle \left|f(x) - 1/3 \right| < \epsilon \).
 
  • #3
Suppose I were to re-construct the solution per your suggestion, how might I go about doing so?
 
  • #4
First you need to find $\delta$. You could solve two systems of inequalities. Suppose that
\[
\left\{
\begin{aligned}
2/(x+3)-1/3&<\epsilon\\
x&<3
\end{aligned}\right.
\]
gives $x>x_1(\epsilon)$ and
\[
\left\{
\begin{aligned}
1/3-2/(x+3)&<\epsilon\\
x&>3
\end{aligned}\right.
\]
gives $x<x_2(\epsilon)$. Then take $\delta_1=3-x_1(\epsilon)$ and $\delta_2=x_2(\epsilon)-3$. Now prove carefully that $x_1(\epsilon)<x<3$ implies $0<2/(x+3)-1/3<\epsilon$ and $3<x<x_2(\epsilon)$ implies $0<1/3-2/(x+3)<\epsilon$. Basically, you need to reverse the process of finding $x_1$ and $x_2$, but this time the proof should start with assumptions and every statement must strictly follow from previous ones. Pay attention to things like multiplying inequalities by negative numbers. Finally, take $\delta=\min(\delta_1,\delta_2)$.

Perhaps there is an easier way that uses some upper or lower bound to simplify the expressions involved.
 
  • #5
Evgeny.Makarov said:
Perhaps there is an easier way that uses some upper or lower bound to simplify the expressions involved.

Indeed, and alternatively, he can suppose $\delta = 1$ and find an upper bound for $|x+3|$.
 
  • #6
Rido12 said:
Indeed, and alternatively, he can suppose $\delta = 1$ and find an upper bound for $|x+3|$.
I am not sure I understand. The value $\delta=1$ does not work for every $\epsilon$.
 
  • #7
I've seen that used commonly in solving quadratic limits. As I understand it, since the concept of limits applies only when $x$ is close to $a$, which in this case is $3$, then we can place a restriction on $x$ such that it is at most $1$ of $a$. (i.e $|x-3|<\delta=1$)
 
  • #8
OK, so you are suggesting finding $\delta$ in the form $\delta=\min(\ldots,1)$. But I still think more information is needed concerning why we need the upper bound on $|x+3|$.
 
  • #9
From the OP's post, we have $|\dfrac{x -3}{3(x + 3)}|<\epsilon$. If we can replace or bound $|x+3|$, then we can easily find delta. I guess, this may also work with a lower bound, but I have never done that. And I have not investigated yet on why it works. Perhaps you can explain. :D
 
  • #10
Rido12 said:
From the OP's post, we have $|\dfrac{x -3}{3(x + 3)}|<\epsilon$. If we can replace or bound $|x+3|$, then we can easily find delta.
The way I understand your suggestion is as follows.

Let $\delta=\min(1,15\epsilon)$. Then $|x-3|<\delta$ implies
\begin{align}
\frac{|x-3|}{3|x+3|}&<\frac{\delta}{3\cdot5}\le\frac{15\epsilon}{15}=\epsilon.
\end{align}
Yes, this is much simpler.
 

FAQ: Epsilon-Delta Proof (Right or Wrong)?

What is an Epsilon-Delta proof?

An Epsilon-Delta proof is a method used in mathematics to prove the limit of a function. It involves using two variables, epsilon (ε) and delta (δ), to show that for any given ε, there exists a δ such that the distance between the input and the output of the function is less than ε.

How is an Epsilon-Delta proof used?

Epsilon-Delta proofs are used to rigorously prove the limit of a function. They are commonly used in calculus and other branches of mathematics to prove the convergence of sequences and series, and to prove the continuity of functions.

What does it mean for an Epsilon-Delta proof to be "right" or "wrong"?

An Epsilon-Delta proof can be considered "right" if it follows the correct logical steps and leads to a correct conclusion. On the other hand, it can be considered "wrong" if there are errors in the reasoning or if the conclusion is incorrect.

What are some common mistakes made in Epsilon-Delta proofs?

Some common mistakes in Epsilon-Delta proofs include using the wrong variables or definitions, not considering all possible cases, and making incorrect assumptions about the function or its limit. It is important to carefully follow the steps of the proof and to double-check all definitions and assumptions.

Are there any tips for successfully completing an Epsilon-Delta proof?

Yes, some tips for successfully completing an Epsilon-Delta proof include carefully reading and understanding the problem, starting with a clear and precise definition of the limit, and breaking down the proof into smaller, more manageable steps. It is also helpful to check the proof for errors and to use examples and counterexamples to test the logic.

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