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I wish to show that for $f:(0,\infty)$, with $f(x) = x^3$, is continuous for all $a\in(0,\infty)$.

I wondered whether the following trick is valid?

Notice that $|x^3-a^3|=|x-a|\cdot|x^2+ax+a^2|$. Now since $a,x$ are strictly positive, surely $x^2+ax+a^2<x^2+2ax+a^2=(x+a)^2$. Then $|x^3-a^3|<|x-a|(x+a)^2<\delta(x+a)^2$. Taking $\delta<1$, we see that $x+a<2a+1$.

Hence given $\epsilon>0$, take $\delta:=\min(1,\frac{\epsilon}{(2a+1)^2})$. Then for all $x\in(0,\infty)$ with $0<|x-a|<\delta$ we have $$|x^3-a^3|<\delta(x+a)^2<\delta(2a+1)^2\leq\epsilon.$$

Have I made a mistake or was this a valid 'trick' in this particular context?

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  • $\begingroup$ You need $|x+a| < |2a+1|$, but otherwise looks good! $\endgroup$ – Dzoooks Nov 16 '19 at 21:02
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    $\begingroup$ @Jack you are correct. Usually I would just use more easily applicable results. However, this question stems from a past-exam question asking us to prove the result in question from first principles. My approach was not included in the suggested solutions, so wished to check that it was correct. $\endgroup$ – Benjamin Nov 16 '19 at 21:45
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    $\begingroup$ Since your argument says for all $x\in(0,\infty)$, you do not need to worry about the case when $x>0$. $\endgroup$ – user9464 Nov 16 '19 at 22:24
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    $\begingroup$ @Dzoooks. $x$ and $a$ are strictly positive because they are two points in the domain $f$ which is $(0, \infty)$. I wasn't basing this one anything to do with $x$ and $a$ being within vicinity of each other. $\endgroup$ – fleablood Nov 16 '19 at 23:06
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    $\begingroup$ @Benjamin Because the domain is strictly positive we know $x,a, x+a, 2a + 1$ are all positive and $|x+1| = x+1 < 2a + 1 = |2a+1|$. ... But it is odd that that is part of the question. If we had had $f:\mathbb R \to \mathbb R$ as $f(x)= x^3$ then $f$ is still continuous everywhere and your trick would still be valid. But then you'd have to show $|x+a| < |2a+1|$. $\endgroup$ – fleablood Nov 16 '19 at 23:13
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  • Since you are proving continuity, instead of $0<|x-a|<\delta$, you need $|x-a|<\delta$ in your proof.

  • all you need is a bound on the quantity $(x+a)^2$ for $x$ near $a$.

  • What you did is OK but I would rewrite it in a slightly different way as follows.


Let $a$ be a fixed real number. Let $\epsilon>0$. We want to find a $\delta>0$ so that for all $x>0$ with $|x-a|<\delta$, $$ |x^3-a^3|<\epsilon.\tag{1} $$

Observe that since $x,a>0$, $$ |x^3-a^3|=|x-a|\cdot|x^2+ax+a^2|=|x-a|(x^2+ax+a^2)\le |x-a|(x+a)^2\tag{2} $$

On the other hand, for $x>0$ with $|x-a|<1$, $$ (x+a)^2=(x-a+a)^2\leq 2(x-a)^2 +2a^2\le 2+2a^2\tag{3} $$ Taking $\delta=\min(1,\frac{\epsilon}{2(1+a^2)})$, we conclude from (2) and (3) that $$ |x^3-a^3|\leq\delta\cdot 2(1+a^2)\leq \epsilon. $$


[Added:] You do not need to write $|x^2+ax+a^2|=x^2+ax+a^2$. Simply use $$ |x^2+ax+a^2|\leq |x|^2+|ax|+a^2 $$ and you can easily bound each of the quantities on the right for $x$ with $|x-a|<1$.

Such approach would give you a proof of continuity of $x^3$ on $\mathbb{R}$.

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If you want to do it from first principles, then using elementary properties of parabolas, note that

for $\delta<a/2,$ we have

$|x^3-a^3|=|x-a|\cdot|x^2+ax+a^2|\le |x-a|\cdot ((a+\delta)^2+a\delta+a^2)\le \frac{21}{4}\cdot |x-a|$

so we may take $\delta =\min(a/2,\epsilon/6).$

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