0
$\begingroup$

I came across this proof here https://en.wikibooks.org/wiki/Calculus/Proofs_of_Some_Basic_Limit_Rules (please see the proof about the Product Rule).

Recall that the definition of a limit is

$$\lim_{x \to a} f(x) =L \iff \forall \epsilon > 0, \exists \delta > 0 : 0 < |x-a| < \delta \implies |f(x) - L| < \epsilon$$

In the product proof they assume/define ${\displaystyle \lim _{x\to c}f(x)=L}$ and ${\displaystyle \lim _{x\to c}g(x)=M}$

With all this in mind, the relevant part I am asking about in the product proof is when they are stating:

Let $\varepsilon$ be any positive number. The assumptions imply the existence of the positive numbers $\delta_{1},\delta_{2},\delta_{3}$ such that

${\displaystyle (1)\qquad {\Big |}f(x)-L{\Big |}<{\frac {\varepsilon }{2(1+|M|)}}}$ when ${\displaystyle 0<|x-c|<\delta _{1}} $

${\displaystyle (2)\qquad {\Big |}g(x)-M{\Big |}<{\frac {\varepsilon }{2(1+|L|)}}}$ when ${\displaystyle 0<|x-c|<\delta _{2}}$

${\displaystyle (3)\qquad {\Big |}g(x)-M{\Big |}<1}$ when ${\displaystyle 0<|x-c|<\delta _{3}}$

Which of these is it saying, equivalently?

  1. $\forall\epsilon>0,\exists\delta_1>0:0<|x-c|<\delta_1\implies|f(x)-L|<\dfrac{\epsilon}{2(1+|M|)}$

  2. $\forall\epsilon>0,\exists\delta_2>0:0<|x-c|<\delta_2\implies|g(x)-M|<\dfrac{\epsilon}{2(1+|L|)}$

  3. $\forall\epsilon>0,\exists\delta_3>0:0<|x-c|<\delta_3\implies|g(x)-M|<1$

Or is it not making these claims for all epsilon and is instead saying

  1. $\exists\delta_1>0:0<|x-c|<\delta_1\implies|f(x)-L|<\dfrac{\epsilon}{2(1+|M|)}$
  2. $\exists\delta_2>0:0<|x-c|<\delta_2\implies|g(x)-M|<\dfrac{\epsilon}{2(1+|L|)}$
  3. $\exists\delta_3>0:0<|x-c|<\delta_3\implies|g(x)-M|<1$

Or, is it nonsensical to state the numbered deltas here, and it is better to say

  1. $\exists\delta>0:0<|x-c|<\delta_1\implies|f(x)-L|<\dfrac{\epsilon}{2(1+|M|)}$
  2. $\exists\delta>0:0<|x-c|<\delta_2\implies|g(x)-M|<\dfrac{\epsilon}{2(1+|L|)}$
  3. $\exists\delta>0:0<|x-c|<\delta_3\implies|g(x)-M|<1$

Or is it without any notation at all:

  1. $\exists\delta>0:0<|x-c|<\delta\implies|f(x)-L|<\dfrac{\epsilon}{2(1+|M|)}$
  2. $\exists\delta>0:0<|x-c|<\delta\implies|g(x)-M|<\dfrac{\epsilon}{2(1+|L|)}$
  3. $\exists\delta>0:0<|x-c|<\delta\implies|g(x)-M|<1$
$\endgroup$
  • $\begingroup$ I gave your formula numbers. This should make it easier to refer to each of them. Change it back, if you don't like that. $\endgroup$ – P. Siehr Mar 7 '18 at 16:58
  • 1
    $\begingroup$ In 10, 11 and 12 you have only removed some of the numbering. $\endgroup$ – Will R Mar 7 '18 at 16:58
  • $\begingroup$ It is saying that for each particular positive $\varepsilon$, the penultimate block 7-9 is true. If you then set $\delta=\min\{\delta_1,\delta_2,\delta_3\}$, you can turn this into a version without subscripts where $\exists\delta>0:\,0<|x-c|<\delta$ implies all three inequalities simultaneously. $\delta$ is likely to be affected by $\varepsilon$ $\endgroup$ – Henry Mar 7 '18 at 16:59
0
$\begingroup$

Formula $(1)$ is a direct application of the definition of a limit: $$\lim_{x \to a} f(x) =L \iff \forall \epsilon > 0, \exists \delta > 0 : 0 < \lvert x-a\rvert < \delta \implies \lvert f(x) - L\rvert < \epsilon.\tag{14.1}$$

Note that in this definition, $\epsilon$ and $\delta$ are both governed by quantifiers. That means the symbols $\epsilon$ and $\delta$ in the definition have no necessary relationship to any symbols that appear anywhere outside the definition. You can replace one or both of these symbols inside the definition, for example, $$\lim_{x \to a} f(x) =L \iff \forall \eta > 0, \exists \delta > 0 : 0 < \lvert x-a\rvert < \delta \implies \lvert f(x) - L\rvert < \eta \tag{14.2}$$ (using $\eta$ instead of $\epsilon$ in this example), and the definition means exactly the same thing as it did before.

In Formula $(1),$ $\epsilon$ is a quantity that will be used further on in the proof. It does not have to be the same $\epsilon$ that was used in your original definition of $\lim_{x \to a} f(x) = L$ (Formula $(14.1)$ above). But $\frac{\epsilon}{2(1+\lvert M\rvert)} > 0,$ so Formula $(14.2)$ says that we can set $$\eta = \frac{\epsilon}{2(1+\lvert M\rvert)},$$ and then (because the rest of the formula is true for any $\eta > 0$), there is some positive number $\delta$ such that $$\lvert f(x)-L\rvert < \eta = \frac{\epsilon}{2(1+\lvert M\rvert)} \text{ whenever $0 < \lvert x-a\rvert < \delta.$}$$ Now that we've established that such a number $\delta$ exists, call it $\delta_1.$ Then we have justified writing Formula $(1)$ exactly as shown in the proof.

In other words, Formula $(1)$ says exactly what it says. The justification for saying what it says is the definition of a limit, where we are careful not to let certain symbols in the definition (which have meaning only inside the definition) "leak out" and confuse us about the meanings of other statements that are not actually part of the definition (even if we want to invoke the definition to justify them).

If you're a programmer, you can think of $\epsilon$ and $\delta$ as local variables inside a context that starts and ends with the start and end of the definition of a limit. The $\epsilon$ in Formula $(1)$ is a variable defined in a different context. It's the same kind of naming logic that allows you to write a function call like y = foo(x/17) when the definition of the function is int foo(int x) { return 2*x; }.

$\endgroup$
  • $\begingroup$ So we could technically just use $\delta$ and $\epsilon$ (no sub-notation) in all three of those assumptions, right? (I added three more cases to the OP). But then in this case how do we designate that we are invoking the deltas and taking the min if we aren't saying which one corresponds where? $\endgroup$ – user538444 Mar 7 '18 at 18:50
0
$\begingroup$

I still have difficulties, to understand your question, but I will try to answer it.


One thing that I learned in my math courses is, that people like to beautify results, which sometimes leads to more difficult understanding.

One of these cases can be seen in this proof. The proof wants to show: $$\lim_{x→a}f(x)g(x) = LM.\label{a}\tag{a}$$

This is, by definition, equivalent to: $$\forall \epsilon > 0, \exists \delta > 0 : 0 < |x-a| < \delta \implies |f(x)g(x) - LM| < ε.\label{b}\tag{b},$$

as you stated in the question.

Now the thing is, that it is considered "beautiful" if you do this proof, do some estimates, and at the final estimate you can write "$<ε$" and not something like "$<5.2ε"$. But it does not matter, since: $$\forall \epsilon > 0, \exists \delta > 0 : 0 < |x-a| < \delta \implies |f(x)g(x) - LM| < cε, \label{c}\tag{c}$$

with some constant $c$ independent of $ε$ is equivalent to \eqref{b}.

Is this part clear?


Let us rewrite some lines of the proof on that wiki-page, under the assumption $|g(x)|<\tilde{c}$. This will get rid of the part about condition (3). If you understand the idea behind the next lines, you can get rid of the simplification as an exercise.

Let $ε>0$ be an arbitrary number. This $ε$ is the $ε$ in \eqref{b}, as we need to show that $∀ε\ ∃δ\ … ⇒ …$

Then by definition of $\lim_{x→a} f(x) = L$ and $\lim_{x→a} g(x)=M$, it holds: \begin{align*} ∃δ_1\ &:\ 0<|x-a|<δ_1\ \ ⇒|f(x) - L|<ε \\ ∃δ_2\ &:\ 0<|x-a|<δ_2\ \ ⇒|g(x) - M|<ε \\ \end{align*}

Why? Because $$\lim_{x→a} f(x) = L ⇔ \forall \tilde{ε} > 0, \exists \tilde{\delta} > 0 : 0 < |x-a| < \tilde{\delta} \implies |f(x) - L| < \tilde{ε}.$$ So especially for the choice $\tilde{ε}:=ε$ there exists such a $\tilde{δ}$. Now name $\tilde{δ}=:δ_1$. Do the same for $g$.

Now if we choose $δ:=\min\{δ_1,δ_2\}$, then both conditions are fulfilled. It holds: \begin{align*} 0<|x-a|<δ\ &\ ⇒|f(x) - L|<ε \\ 0<|x-a|<δ\ &\ ⇒|g(x) - M|<ε \\ \end{align*}

Keep in mind that $δ_1$ and $δ_2$ simply give us intervals on the x-axis around the point $a$ with diameter $δ_{1|2}$. So we just chose the smaller interval.

Then we can finish the proof with: \begin{align*} |f(x)g(x) -LM| &= |fg-Lg+Lg-LM| \\ &≤|fg-Lg|+|Lg-LM| \\ &=|g||f-L|+|L||g-M| \\ &≤|g|ε+|L|ε \\ &≤ \underbrace{(\tilde{c}+|L|)}_{=:c}ε \end{align*}

What have we done here: We chose an arbitrary $ε>0$ $[\hat{=}∀ε]$ found a $δ$ $[\hat{=}∃δ]$, such that $$0<|x-a|<δ ⇒ |f(x)g(x) - LM| < cε.$$

I hope this makes the proof understandable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.