# Proving that $\lim_{x \to a}f(x)=\lim_{h\to 0}f(a+h)$: How is a change in variable implemented in a universally quantified statement.

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, $$\color{red}{\text{substitute h+a for x}}$$:

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

Simplifying:

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

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 $$h$$. $$\quad \square$$

Question

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.

• 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. – S.Cramer Apr 14 at 2:28
• @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. – Brian M. Scott Apr 14 at 2:33
• @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. – Brian M. Scott Apr 14 at 2:40
• I'll mull over it a little bit more, but I think I understand the distinction. Cheers, Sir~ – S.Cramer Apr 14 at 2:48
• 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. – 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 $$\phi(t)$$.