MHB Epsilon-Delta Proof (Right or Wrong)?

[bwpbruce]

Given:
[math]\lim_{x \to 3} \dfrac{2}{x + 3} = \dfrac{1}{3}[/math]

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

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

Solve:
[math] \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 [/math]
\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:
[math] 18\epsilon = \delta [/math]

Conclusion:
From the statement [math]18\epsilon = \delta [/math], it is clear that for any given [math]\epsilon[/math], a [math]\delta[/math] exists such that if [math] 0 < \left|x - 3 \right| < \delta [/math], then [math] \left|f(x) - \dfrac{1}{3} \right| < \epsilon [/math]

 

[Evgeny.Makarov]

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 [math] \left|f(x) - 1/3 \right| < \epsilon [/math]), 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:

\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 [math] \left|f(x) - 1/3 \right| < \epsilon [/math].

 

[bwpbruce]

Suppose I were to re-construct the solution per your suggestion, how might I go about doing so?

 

[Evgeny.Makarov]

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.

 

[Dethrone]

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 $|x+3|$.

 

[Evgeny.Makarov]

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 $\epsilon$.

 

[Dethrone]

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 $|x-3|<\delta=1$)

 

[Evgeny.Makarov]

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 $|x+3|$.

 

[Dethrone]

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

 

[Evgeny.Makarov]

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.