# Is this trick valid in proving that $f:(0,\infty)\to\mathbb{R}$; $f(x)=x^3$ is continuous?

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 $$x^2+ax+a^2. Then $$|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 $$|x^3-a^3|<\delta(x+a)^2<\delta(2a+1)^2\leq\epsilon.$$

• You need $|x+a| < |2a+1|$, but otherwise looks good! – Dzoooks Nov 16 '19 at 21:02
• @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. – Benjamin Nov 16 '19 at 21:45
• Since your argument says for all $x\in(0,\infty)$, you do not need to worry about the case when $x>0$. – user9464 Nov 16 '19 at 22:24
• @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. – fleablood Nov 16 '19 at 23:06
• @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 $|x+a| < |2a+1|$. – fleablood Nov 16 '19 at 23:13

• 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 $$\mathbb{R}$$.

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

for $$\delta 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 $$\delta =\min(a/2,\epsilon/6).$$