Ε-δ proof: lim x->a f(x) = lim h->0 f(a + h)

[MeyCey]

This is a simple exercise from Spivak and I would like to make sure that my proof is sufficient as the proof given by Spivak is much longer and more elaborate.

Homework Statement

Prove that $$\lim_{x\to a} f(x) = \lim_{h\to 0} f(a + h)$$

Homework Equations

The Attempt at a Solution

By the definition of limit (omitting the 'for all's and 'there exists''):
$$\lim_{x\to a} f(x) = L_1 \implies 0 < |x - a| < δ_1 \implies |f(x) - L_1| < ε$$
And
$$\lim_{h\to 0} f(a + h) = L_2 \implies 0 < |h - 0| = |h| < δ_2 \implies |f(a + h) - L_2| < ε$$
Now let's work on the second limit:
$$|h| = |(a+h) - a|\\ \text{Let } y = a + h \\$$Now the second limit takes the following definition:$$0 < |y - a| < δ_2 \implies |f(y) - L_2| < ε \\$$We see that both limits have the exact same form right now. By Theorem 1 [stating that if limit L exists it is necessarily unique] $$L_1 = L_2\\ \text{ and }\\ \lim_{x\to a} f(x) = \lim_{h\to 0} f(a + h)\\ \text{Q.E.D.}$$

Is my proof correct or have I perhaps made a mistake somewhere?

 

[Stephen Tashi]

This is a simple exercise from Spivak and I would like to make sure that my proof is sufficient as the proof given by Spivak is much longer and more elaborate.

3. The Attempt at a Solution

By the definition of limit (omitting the 'for all's and 'there exists''):

Omitting the quantifiers renders your proof insufficient. Of course, you might be taking course where the standard for doing a proof in homework is to write down a series of symbolic manipulations. So whether your work meets the standard of given instructor for homework is going to depend on the instructor.

What you wrote could guide an experienced mathematician to write a proof. That proof would be given using complete sentences. (Coherent mathematical abbreviations can form components of complete sentences. )

 

[PeroK]

Let me guess what Spivak does:

a) Doesn't assume that both limits exist and are real numbers.

b) Shows that if one limit exists (and is a real number), then the other exists and is the same real number.

c) Shows that if one limit is $\pm \infty$ then so is the other.

 

[MeyCey]

Omitting the quantifiers renders your proof insufficient.

Yes, I understand that the proof is not complete; I omitted the quantifiers for the sake of brevity. My question was intended to be more in line of 'Is my reasoning correct?' than 'Is the proof a complete, rigorous proof?', i.e. is all I have to do from that point on is add the correct quantifiers, etc., though I now see how it was ambiguous.

As for the Spivak's proof it seems to be quite similar to my reasoning after I understood the material a bit better--he did also make the assumption of the limits existence. I think that it's because up to the point he didn't define what it means for a limit to be equal to $\pm \infty$ (he defined it in a later exercise) -- so such a limit would be deemed as non-existing.

 

[Stephen Tashi]

Yes, I understand that the proof is not complete; I omitted the quantifiers for the sake of brevity. My question was intended to be more in line of 'Is my reasoning correct?' than 'Is the proof a complete, rigorous proof?', i.e. is all I have to do from that point on is add the correct quantifiers, etc., though I now see how it was ambiguous.

One thing to fix-up about your abbreviations is that mathematical definitions are logical equivalences and somewhere in your outline, you need to use the fact that they are. ( ""$\iff$"" instead of "$\implies$").

As for the Spivak's proof it seems to be quite similar to my reasoning after I understood the material a bit better--he did also make the assumption of the limits existence. I think that it's because up to the point he didn't define what it means for a limit to be equal to $\pm \infty$ (he defined it in a later exercise) -- so such a limit would be deemed as non-existing.

Yes, in spite of the fact that mathematics attempts to be precise, ambiguity creeps in. Several different types of limits are defined and often we have to infer which type is meant from the context. $lim_{x \rightarrow a} f(x) = L$ defines one type of limit when $L$ is a number. Different types are defined when the notation changes to allow "$a$" or "$L$" to be replaced by "$\infty$" or "$-\infty$".

Distinguishing among these 9 different types of limits sometimes requires attention to the fine print. For example, if a theorem begins by saying "If $\lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} g(x)$ then...", we have to worry about whether this is a hypothesis that applies when both limits are of a type involving "$\infty$". Sometimes the hypothesis "If $\lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} g(x)$..." is intended to include the case when both limits fail to exist or fail to exist as limits of one type by existing as limits of another type.

 

[MeyCey]

One thing to fix-up about your abbreviations is that mathematical definitions are logical equivalences and somewhere in your outline, you need to use the fact that they are. ( ""$\iff$"" instead of "$\implies$").

Oh, so the first line of a complete proof should read:
$$\lim_{x\to a} f(x) = L_1 \iff \forall ε > 0 \ \exists δ_1 > 0 \ : \ 0 < |x - a| < δ_1 \implies |f(x) - L_1| < ε$$
And the rest would have to be corrected analogously. Is it correct now, or am I still missing something?

 

[Stephen Tashi]

And the rest would have to be corrected analogously. Is it correct now, or am I still missing something?

I can't tell if you are missing anything until you give an actual proof.

 

[MeyCey]

I can't tell if you are missing anything until you give an actual proof.

Ok, so the full proof would read:

By the definition of limit:
$$\lim_{x\to a} f(x) = L_1 \iff \forall ε > 0 \ \exists δ_1 > 0 \ : \ 0 < |x - a| < δ_1 \implies |f(x) - L_1| < ε$$
And
$$\lim_{h\to 0} f(a + h) = L_2 \iff \forall ε > 0 \ \exists δ_2 > 0 \ : \ 0 < |h - 0| = |h| < δ_2 \implies |f(a + h) - L_2| < ε$$
By manipulating the second limit we get:
$$|h| = |(a+h) - a|\\ \text{Let } y = a + h \\$$Now, the second limit takes the following form:$$\forall ε > 0 \ \exists δ_2 > 0 \ : \ 0 < |y - a| < δ_2 \implies |f(y) - L_2| < ε \\$$We see that both limits have the exact same form right now. By Theorem 1 [stating that if limit L exists it is necessarily unique] $$L_1 = L_2\\ \text{ and }\\ \lim_{x\to a} f(x) = \lim_{h\to 0} f(a + h)\\ \text{Q.E.D.}$$

That is meant to be full proof with all the formalities respected. Is there any flaw in it?

 

[Stephen Tashi]

You have the correct idea and some people would accept your work as an informal proof. However, your statement that "both limits have the exact same form" is not a precisely defined concept. Each of $L_1, L_2$ is technically a number and the goal is to prove that two numbers are equal. So what would we mean by saying that two numbers have "the exact same form"?

A mathematical definition can (with some effort!) be expressed in the "form": [expression containing undefined things] $\iff$ [statement with a previously defined meaning]. In the definition of $lim_{x \rightarrow a} f(x) = L$ the whole expression and notation "$lim_{x \rightarrow a} f(x) = L_1$" is the undefined thing. The statement with the previously defined meaning is "$\forall \epsilon > 0,\ \exists \delta > 0: 0 < |x-a| < \delta \implies |f(x) - L| < \epsilon$".

Included in all that notation is a lot of "cultural" understanding that is not verbally expressed - for example, we understand that "$f$" is a real valued function of one real variable and that the other variable mentioned refer to real numbers.

To state your concept precisely, you need to argue that the number $L_1$ is the unique number that statisfies a certain logical statement, that $L_2$ is the unique number that satisfies a second logical statement, and that the two logical statements are logically equivalent. Depending on how critical your audience is about observing the formalities of proving statements logically equivalent, you might get by with arguing that two logical statements are logically equivalent because" they have the same form" - or you might be required to show two statements are logically equivalent the old fashioned way by proving that each statement implies the other.