Correction of a simple continuity problem I just learned the definition of continuity. What I want to show is that if $f(x)=x$ is continuous at $a$, then $g(x)=x^3$ is continuous at $a$ and has the limit $a^3$ (functions from $\Bbb R\to \Bbb R$)
I proceeded as follow: For any $\epsilon >0$, let's take $T = \min(a - \sqrt[3]{a^3-\epsilon};\sqrt[3]{a^3+\epsilon}-a)$. Then, $-T+a\le x \le T+a$ implies $|x^3 - a^3| \le e$ thus $g$ is continuous at $a$...
I Did the following to find $T$:
Suppose that for every $\epsilon>0$, 
$$\sqrt[3]{a^3 - \epsilon} \le x \le\sqrt[3]{a^3 +\epsilon}$$
Let's take $T$ such that $\sqrt[3]{a^3 -\epsilon} \le -T+a \le x \le T+a \le \sqrt[3]{a^3+\epsilon}$, and I had the idea to take$T = \min(a - \sqrt[3]{a^3-\epsilon};\sqrt[3]{a^3+\epsilon}-a)$ because it works, but well, it's not really beautiful, that's so ugly,..
Is there any other idea?
 A: First of all, given the way that the problem is worded, I do not think that you were intended to approach this with an $\epsilon,\delta$ proof. Instead you are intended to use the symbolically cleaner definition that a function $f$ is continuous at $a$ if $$\lim_{x\to a} f(x) = f(a)$$
I believe this because you are given that $f(x) = x$ is continuous at $a$. This is almost obvious. Only the proof of continuity for constant functions is simpler. In proving continuity for $g(x) = x^3$ by $\epsilon,\delta$, you don't need it. So why is it there?
I would bet that somewhere in your course, it was already proven about limits that for any two functions $g, f$, if both $\displaystyle\lim_{x \to a} f(x)$ and $\displaystyle\lim_{x \to a} g(x)$ converge, then so does $\displaystyle\lim_{x \to a} f(x)g(x)$ and
 $$\lim_{x \to a} f(x)g(x) = \left(\lim_{x \to a} f(x)\right)\left(\lim_{x \to a} g(x)\right)$$
Therefore, what you are intended to do here is recognize that for the particular functions $f$ and $g$ given, that $g(x) = f(x)f(x)f(x)$. Then apply the multiplication of limits rule above to see that 
$$\lim_{x \to a} g(x) = \left(\lim_{x \to a} f(x)\right)\left(\lim_{x \to a} f(x)\right)\left(\lim_{x \to a} f(x)\right)$$
However, Your proof is also correct, and I'm afraid that $\epsilon,\delta$ proofs are not known for being beautiful. (Well - they are, but it is a beauty that takes much experience to recognize. It is like coffee. Most people find it revolting at first taste, but those who keep trying it learn to appreciate the subtleties of its flavor.)
A: Here is one $\ \delta\,$ that looks a little more "beautiful"
If $\ a\neq0\ $then set 
$$0\,<\,\delta\,<\,\min\left(\frac{\varepsilon}{8a^2},|a|\right)$$
First note that $\ |x-a|<\delta\ \Rightarrow\ |x|<\delta+|a|<2|a|$,$\ \,$then we have 
$$|x^3-a^3|\,=\,|x-a|\cdot|x^2+ax+a^2|\,<\,\delta\,(|4a^2|+|2a^2|+|a^2|)\,<\,\frac{\varepsilon}{8a^2}\cdot7a^2\,<\,\epsilon$$
If $\ a=0\ $ then set 
$$0\,<\,\delta\,<\,\sqrt[3]{\varepsilon}$$
and the proof is trivial

$\left.\right.$
We can use the same idea to prove the continuity of $\ f(x)=x^n\,$ at $\ a\ $where $\,n\,$ is an positive integer.
Similarly, if $\ a\neq0\ $then set
$$0\,<\,\delta\,<\,\min\left(\frac{\varepsilon}{2^n|a^{n-1}|},|a|\right)$$
Hence we have 
$$|x^n-a^n|=|x-a||x^{n-1}+x^{n-2}a+\cdots+xa^{n-2}+a^{n-1}|$$
$$\qquad\qquad\qquad\quad\,<\ \delta\,(\,2^{n-1}|a|^{n-1}+2^{n-2}|a^{n-1}|+\cdots+2|a^{n-1}|+|a^{n-1}|\,)$$
$$=\ \delta(2^{n}-1)|a^{n-1}|\qquad\qquad\qquad$$
$$<\ \frac{\varepsilon}{2^n|a^{n-1}|}(2^{n}-1)|a^{n-1}|\qquad\quad\ $$
$$<\ \varepsilon\qquad\qquad\qquad\qquad\qquad\quad\ $$
Then for $\ a=0$, $\ $simply set
$$0\,<\,\delta\,<\,\sqrt[n]{\varepsilon}$$
