I am learning calculus more rigorously using Spivak's Calculus (4th ed.) and I stumbled upon a question I would like some help with. Consider this simple example: Let us show that
$$ \lim_{x\to a}x^3 =a^3. $$
Using a direct approach (using the definition of a limit), I am able to show that $\delta=\min\left(1,\dfrac{\epsilon}{3|a|^2+3|a|+1}\right)$ is a suitable choice to do an $\epsilon \delta$ proof (assuming I did everything correctly). For functions that are built out of simpler functions (e.g. $x^3, 2x^2-1$, etc.), ch.5 pages 102-105 of the book (relevant info at the end of post) show us that we can not only find the limit but also what would be a suitable $\delta$ to prove the limit of the function in question.
For $x^3$, the proof of this theorem suggests that we can find a suitable $\delta$ by letting $f(x) = x$ and $g(x) = x^2$ so that knowing that $f \to a, g\to a^2$ as $x\to a$, it is trivial to conclude that $f\cdot g \to a^3$. We also know that $\delta = \epsilon$ and $\delta = \min(1,\epsilon/(2|a|+1))$ are suitable choices for $f$ and $g$ respectively. Now, to find a $\delta$ using this prescription, we have
$$ \begin{align} \text{if} \; 0 &<|x-a|<\delta_1, \text{then} \; |x - a| < \min\left(1,\dfrac{\epsilon}{2(|a|^2+1)}\right), \\ \text{and} \quad \text{if} \; 0 &<|x-a|<\delta_2, \text{then} \; |x^2 - a^2| <\dfrac{\epsilon}{2(|a|+1)}. \end{align} $$
Using the $\delta's$ known to work for $f$ and $g$, I have
$$ \begin{align} \delta_1 &= \min\left(1,\dfrac{\epsilon}{2(|a|+1)} \right), \\ \delta_2 &= \min\left(1,\dfrac{\frac{\epsilon}{2(|a|^2+1)}}{2|a|+1} \right) = \min\left(1,\dfrac{\epsilon}{2(|a|^2+1)(2|a|+1)} \right) \end{align} $$
and therefore $\delta = \min(\delta_1,\delta_2)$. However at a superficial glance, this does not seem to give me the same delta obtained via direct approach. So here are my questions:
- How can I relate these two $\delta's$ I found? Will they be the same? I know that for $\epsilon\delta$ proofs, they are not unique, but I was hoping I could use this method to double check my work.
- Could someone help me out untangling these $\min$ expressions? To see what comes after the dust settles?
Thank you for your time and help.
For those that do not have the book, here are the relevant facts:
LEMMA (without proof):
If $|x-x_0|<\min\left(1,\dfrac{\epsilon}{2(|y_0|+1)} \right)$ and $|y-y_0|<\dfrac{\epsilon}{2(|x_0|+1)}$, then $|xy-x_0y_0|<\epsilon$.
THEOREM (Only the relevant part for this question)
If $\lim_{x\to a}f(x) = l$ and $\lim_{x\to a}g(x) = m$, then $\lim_{x\to a}(f\cdot g)(x) = l\cdot m$, where the dot indicates regular multiplication.
I will need the proof of this theorem, so I will quote it as well. The hypothesis means that $\forall \epsilon>0$ there are $\delta_1,\delta_2>0$ such that
$$ \begin{align} \text{if} \; 0 &<|x-a|<\delta_1, \text{then} \; |f(x) - l| <\epsilon, \\ \text{and} \quad \text{if} \; 0 &<|x-a|<\delta_2, \text{then} \; |g(x) - m| <\epsilon. \end{align} $$
This means that we can consider
$$ \begin{align} \text{if} \; 0 &<|x-a|<\delta_1, \text{then} \; |f(x) - l| <\min\left(1,\dfrac{\epsilon}{2(|m|+1)}\right), \\ \text{and} \quad \text{if} \; 0 &<|x-a|<\delta_2, \text{then} \; |g(x) - m| <\dfrac{\epsilon}{2(|l|+1)}. \end{align} $$
where we have consulted the lemma for a suitable choice form of $\epsilon$. Let $\delta = \min(\delta_1,\delta_2)$. If $0<|x-a|<\delta$, then by the lemma it follows that $|f\cdot g - l\cdot m|<\epsilon$, which proves the theorem.
