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~
Best Answer
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.