# Question on the Validity of $\lim_{x\to a} {x^3} = {a^3}$ Proof from Spivak's Calculus

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 $$a$$?

• 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. – Brian Borchers Jul 22 at 1:16
• 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. – Ethan Bolker Jul 22 at 1:18
• 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 – dm027 Jul 22 at 1:29
• perhaps I'm missing something obvious but how did you get the 2nd line "$\dots < (2|a|+1)^2$"? – peek-a-boo Jul 22 at 1:34
• 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 $\varepsilon$. – peek-a-boo Jul 22 at 1:42

## 1 Answer

Once again, it doesn't matter that you arrive at $$\epsilon$$ precisely!

If for $$|x-a|<\delta$$ you can show that $$|x^3-a^3| with $$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 $$\delta=\min(\epsilon,|a|)\implies |x^3-a^3|<7a^2\epsilon$$.

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