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I was reading through the limits chapter, when I stumbled upon the proof that $f(x) = x^3$ approaches $a^3$ as $x$ approaches $a$

If $$|x - a| < \min \left(1, \frac{\epsilon}{(1+|a|)^2 + |a|(1+|a|) + |a|^2} \right)$$ then $|x^3 - a^3|$ $ < \epsilon$.

This is what Spivak did

If $|(x-a)| < 1$, then $|x| < |a| + 1$ and consequently $$|x^2 + ax + a^2| \leq |x|^2 + |a||x| + |a|^2 < (1+|a|)^2 + |a|(1+|a|) + |a|^2 $$

Therefore \begin{align} |x^3 - a^3| &= |x-a||x^2 + ax + a^2| \\ &< \frac{\epsilon}{(1+|a|)^2 + |a|(1+|a|) + |a|^2}(1+|a|)^2 + |a|(1+|a|) + |a|^2 \\ &= \epsilon \end{align}

I tried doing it on my own before seeing the result and I got

\begin{align} |x^2 + ax + a^2| &\leq |x|^2 + |a||x| + |a|^2 \\ &< (2|a| + 1)^2 \\ &= 4a^2 + 4|a| + 1 \end{align}

which is almost the same since $(1+|a|)^2 + |a|(1+|a|) + |a|^2 = 3a^2 + 3|a| + 1$.

So I was wondering if my answer is invalid or not, and if not, does that mean some limits have several possible $\delta$ values when $x$ approaches some $a$?

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    $\begingroup$ If $\delta$ is sufficient for a given $\epsilon$, then $\delta/2$, $\delta/3$, and infinitely many other values will also work! For proving the limit we don't care about getting the largest possible $\delta$, so whatever makes your proof simple is good enough. $\endgroup$ – Brian Borchers Jul 22 at 1:16
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    $\begingroup$ I haven't checked your calculations, but the answer to your last question is "yes". If some value of $\delta$ works then any smaller value will too. There's rarely any reason to look for as large a $\delta$ as possible. Often some standard argument or inequality leads to one that will do. $\endgroup$ – Ethan Bolker Jul 22 at 1:18
  • $\begingroup$ I was thinking about that, but I was kind of skeptical about having a different answer from the one on the book haha, thanks for the insight $\endgroup$ – dm027 Jul 22 at 1:29
  • $\begingroup$ perhaps I'm missing something obvious but how did you get the 2nd line "$ \dots < (2|a|+1)^2$"? $\endgroup$ – peek-a-boo Jul 22 at 1:34
  • $\begingroup$ never mind, I see it now. But yea, so if you choose $\delta = \min \left( 1, \dfrac{\varepsilon}{(2|a|+1)^2}\right)$, then it will also work. Also, like the other comments said, there's often no need to find the "largest" possible $\delta$ (sometimes, there might not even be a largest one). All that matters is you find a particular one which works for the initially given value of $\varepsilon$. $\endgroup$ – peek-a-boo Jul 22 at 1:42
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Once again, it doesn't matter that you arrive at $\epsilon$ precisely!

If for $|x-a|<\delta$ you can show that $|x^3-a^3|<K\epsilon$ with $K$ a constant (i.e. independent of $x$) then it's a win.

Cutting epsilons is just a matter of aesthetic and IMHO just confuse the beginner. This practice is a bit outdated (we nowadays make use of equivalents much more than in the past), and it is not mandatory to have to go for the finest inequalities (i.e. $K=1$) in delta-epsilon proofs, you are just as fine with $k=6446595379f(a)$ or any other value whatsoever...

So you get $|x-a|<\delta$.

For the convergence part you need $\delta<\epsilon$

For the bounded part, you need $|x|<|a|+\delta$ so just do it VERY roughly and choose $\delta<|a|$ (why? well, because it simplifies all calculations).

Then you get $|x|<2|a|$ and $|x^2+ax+a^2|<4a^2+2a^2+a^2=7a^2$

$K=7a^2$ is constant, that is all you need.

Conclude by taking $\delta=\min(\epsilon,|a|)\implies |x^3-a^3|<7a^2\epsilon$.

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  • $\begingroup$ Where were you in 1969 when I first tackled these problems? $\endgroup$ – steven gregory Jul 22 at 2:09
  • $\begingroup$ Thank you! I'll have it in mind when working through the book $\endgroup$ – dm027 Jul 22 at 2:20

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