To provide the appropriate context, I will state the problem in the text book, provide my solution, and then ask my question.

Chapter 5: Problem 9 - Spivak's Calculus

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

Assert that $\displaystyle\lim_{x \to a}f(x)=L$.

By definition, this means that: $\forall \epsilon \gt 0 \quad \exists \delta \gt 0 \quad \forall x \in \mathbb R \big [ 0 \lt \lvert x -a \rvert \lt \delta \rightarrow \lvert f(x) - L \rvert \lt \epsilon\big ]$.

Let's choose an arbitrary instance of this, using an $\epsilon'$ and a corresponding $\delta '$.

$$0 \lt \lvert x -a \rvert \lt \delta' \rightarrow \lvert f(x) - L \rvert \lt \epsilon' \quad \dagger$$

In order to prove that $\displaystyle\lim_{h\to 0}f(a+h)=L$, we need to show that the aforementioned definition is satisfiable. For convenience, define $g(h)=f(a+h)$. (Note that $a$ is a constant). We are therefore proving the following statement:

$$\forall \epsilon \gt 0 \quad \exists \delta \gt 0 \quad \forall h \in \mathbb R \big [ 0 \lt \lvert h -0 \rvert \lt \delta \rightarrow \lvert g(h) - L \rvert \lt \epsilon\big ]$$

In the $\dagger$ statement, $\color{red}{\text{substitute $h+a$ for $x$}}$:

$$0 \lt \lvert (h+a) - a \rvert \lt \delta' \rightarrow \lvert f(h+a) - L \rvert \lt \epsilon'$$


$$ 0 \lt \lvert h \rvert \lt \delta' \rightarrow \lvert g(h) - L \rvert \lt \epsilon' $$

Noting that $\lvert h \rvert = \lvert h - 0 \rvert$, we have proven that we can construct a $\delta$ for an arbitrary $\epsilon$ in which the implication is true.

This proves that statement:

$$\displaystyle\lim_{x \to a}f(x)=L \rightarrow \displaystyle\lim_{h\to 0}f(a+h)=L$$

To demonstrate that:

$$\displaystyle\lim_{h\to 0}f(a+h)=L \rightarrow \displaystyle\lim_{x \to a}f(x)=L $$

work the opposite direction, where the substitution is now $(x-a)$ for $h$. $\quad \square$


I am having trouble understanding what exactly I am doing when I carry out the step, "...substitute $h+a$ for $x$" (and later on when we substitute $x-a$ for $h$).

In fact, I am not even sure I know how to ask the question hah. It appears as though I am saying $x=h+a$, but what does this mean in the context of the logical quantifiers? When I assert $x=h+a$, because $a$ is a constant, I am saying $x=T(h)=h+a$, where $T$ is a function.

Using the above terms, am I now claiming that $\forall x\ \varphi (x) \iff \forall T(h)\ \varphi\big(T(h)\big) \iff \forall (h+a) \ \varphi \big( (h+a) \big)$?

In English, "Any statement about $x$ is equivalent to any statement about $h+a$".

Is this somewhere in the right ball park? Cheers~


You don’t have to make a change of variable.

Assume that $\lim\limits_{x\to a}f(x)=L$, and let $\epsilon>0$; then there is a $\delta_\epsilon>0$ such that $|f(x)-L|<\epsilon$ whenever $0<|x-a|<\delta_\epsilon$. Now suppose that $0<|h|<\delta_\epsilon$; then $0<|(a+h)-a|<\delta_\epsilon$, so $|f(a+h)-L|<\epsilon$, so $\lim\limits_{h\to 0}f(a+h)=L$. The other direction is similar.

  • $\begingroup$ Hmmm. I'm a little confused. Aren't you implicitly setting $x=a+h$? i.e. You are working forward from the claim $\forall \epsilon \gt 0 \quad \exists \delta \gt 0 \quad \forall x \in \mathbb R \big [ 0 \lt \lvert x -a \rvert \lt \delta \rightarrow \lvert f(x) - L \rvert \lt \epsilon\big ]$, and choosing a particular $x$ that satisfies the antecedent. Namely, $x=a+h$. Or are you saying that the statement $x=a+h$ is not an example of a change of variable. $\endgroup$ – S.Cramer Apr 14 at 2:28
  • $\begingroup$ @S.Cramer: Not really. I’m just observing that $a+h$ satisfies the condition tha ensures that $|f(a+h)-L|<\epsilon$. In a sense there really is no $x$: $x$ is a bound variable, not the name of a specific quantity. It’s just a label that lets you talk about the same quantity in more than one place in the first-order formula. $\endgroup$ – Brian M. Scott Apr 14 at 2:33
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    $\begingroup$ @S.Cramer: Any $u$-substitution to compute an integral would qualify. Or suppose that you’re asked to determine the number of solutions in positive integers to $\sum_{k=1}^nx_k=m$; you can instead let $y_k=x_k-1$ for $k=1,\ldots,n$ and determine the number of solutions in non-negative integers to $\sum_{k=1}^ny_k=m-n$ and get the same result, and in so doing you’ve made $m$ changes of variable. $\endgroup$ – Brian M. Scott Apr 14 at 2:40
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    $\begingroup$ I'll mull over it a little bit more, but I think I understand the distinction. Cheers, Sir~ $\endgroup$ – S.Cramer Apr 14 at 2:48
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    $\begingroup$ Feel free to ignore this :) ... just making some documentation for my own purposes. When you specify a "particular" $\epsilon$ and accompanying $\delta_{\epsilon}$, we can assert that $\forall x \big [0 \lt \lvert x -a \rvert \lt \delta_{\epsilon} \rightarrow \lvert f(x) - L \rvert \lt \epsilon \big ]$. Now, we simply choose a value that satisfies the antecedent of this universally quantified implication. The value we choose is $a+h$. The consequent follows. $\endgroup$ – S.Cramer Apr 14 at 3:32

Let's consider statements of the form $\forall x \in \mathbb R [\phi(x)]$. Here, $x$ is a formal variable which exists somewhere in the body of $\phi(x)$. Syntactically speaking, I could replace every occurence of $x$ in the body of $\phi(x)$ with a real number, say $1.3$, and get a different logical formula. Let's consider the specific case where $\phi(x)$ is the expression $x < 100$. I'll get back to the actual expression you care about in a moment. In this case, that replacement yields $1.3 < 100$. I'll denote this new formula by $\phi(x:=1.3)$.

This was a purely syntactic construction - I just replaced every instance of the string "$x$" with the string "$1.3$". But this also has semantic meaning. On its own, $x$ has no meaning. It is not any particular real number, rather, it is a formal variable meant to stand in place for any particular real number. When I consider the formula $\phi(x:=1.3)$, that is an honest proposition I can ask the truth value of. In the example I gave, it evaluates as a true expression. On the other hand, an expression of the form $\phi(x)$ with this free variable $x$ doesn't have a truth value. Is $x < 100$ true? It doesn't even make sense to ask that, as $x$ is not any particular real number.

However, the expression $\forall x \in \mathbb R [\phi(x)]$ is a sensible expression we can talk about the truthhood or falsity of. When is this expression true? Precisely when each and every replacement $\phi(x:=c)$ is true for every real number $c$. So saying that $\forall x[\phi(x)]$ is true implies that that $\phi(x:=1.3)$, $\phi(x:=\pi)$, $\phi(x:=10000000)$, etc. are all true. Of course, $10000000 < 100$ is false, so $\forall x \in \mathbb R [\phi(x)]$ is false in this case.

Getting back to the question at hand, you're interested in $0 \lt \lvert x -a \rvert \lt \delta \rightarrow \lvert f(x) - L \rvert \lt \epsilon$, which I'll again denote $\phi(x)$. But wait, does $\phi(x:=5)$ have meaning? If we replace it, we get $0 \lt \lvert 5 - a \rvert \lt \delta \rightarrow \lvert f(5) - L \rvert \lt \epsilon$. We are given $a$ and $L$, but what are these $\delta$ and $\varepsilon$ symbols? It's meaningless to say whether this, in isolation, is true or false. This is something I brushed past above - you need to quantify (or replace) all free variables in an expression to be able to tell if it's true or false.

Now, when your book said to replace $x$ with $h+a$, here's how I'd interpret it. You have your formula $\psi(h)$ (I'm suppressing the other variables for simplicity) and you want to prove $\forall h\in \mathbb R[\psi(h)]$. By definition, that means that for every real number $c$, $\psi(h:=c)$ is true. Furthermore, you know that the expression $\forall x \in \mathbb R [\phi(x)]$ is true, so in particular the substitution $\phi(x:=c+a)$ is true. You then use this substitution to prove the substitution you're really after, $\psi(x:=c)$ is true. Your book is using a (very common) notational trick where $h$ is treated both as a formal variable and whatever particular value we want it to be at the moment. So when they say to substitute $x$ with $a + h$, they mean that we replace instances of $h$ with a real number $c$ in $\psi$ then we subsequently replace instances of $x$ with the real number $a + c$ in $\phi$. They're avoiding using this extra $c$ and calling it $h$ instead.


Since your question seems to be about logic, I will try to answer from a logician's perspective.

In your statement (with explicit quantifiers), $$\forall x.\forall\epsilon.\exists \delta.\quad 0 \lt \lvert x -a \rvert \lt \delta \rightarrow \lvert f(x) - L \rvert \lt \epsilon \tag{$\diamondsuit$}$$

when you say "substitute $h+a$ for $x$", what you are really saying is

"Let us instead prove $$\forall h.\forall\epsilon.\exists \delta.\quad 0 \lt \lvert (h+a) -a \rvert \lt \delta \rightarrow \lvert f(h+a) - L \rvert \lt \epsilon,\tag{$\spadesuit$}$$ because $\spadesuit$ implies $\diamondsuit$."

Then, to prove that $\spadesuit$ implies $\diamondsuit$, you first introduce a new universally quantified variable $x$ to $\spadesuit$ (this is called weakening): $$\forall x.\forall h.\forall\epsilon.\exists \delta.\quad 0 \lt \lvert (h+a) -a \rvert \lt \delta \rightarrow \lvert f(h+a) - L \rvert \lt \epsilon,\tag{$\spadesuit'$}$$

and then you instantiate $h := (x - a)$ in $\spadesuit'$ (this is called elimination for the universal quantifier): $$\forall x.\forall\epsilon.\exists \delta.\quad 0 \lt \lvert ((x - a) + a) -a \rvert \lt \delta \rightarrow \lvert f((x - a) + a) - L \rvert \lt \epsilon\tag{$\diamondsuit$}$$

So in reality you are not really substituting, but instantiating. The substitution (or Leibniz definition of equality) rule says that whenever $s = t$, then for any proposition $\phi(x)$, then $\phi(s)$ implies $\phi(t)$ (Note that when I say "for any proposition" here this is a meta universal quantification, that is outside first order logic because I am quantifying over propositions). In summary, "substitute $h + a$ for $x$" is a linguistic shortcut, or abuse of language which may mean different things in different contexts.

To get back to your main question I must point out (as others have) that $\spadesuit$ is insufficient to prove your claim, since you have not yet shown how to construct $\delta$. In particular, the introduction rule for existential quantifiers says that in order to prove $\exists x.\phi(x)$, you must construct a $t$ (without using $x$) such that $\phi(t)$.


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