<s>[INST] Proving $\lim_{x\to c}x^3=c^3$ for any $c\in\mathbb R$ using $\epsilon$-$\delta$ definition 
$\lim_{x\to 3}x^3=c^3$ for any $c\in\mathbb R$

Let $\epsilon>0$. Then
$$|x^3-c^3|=|x-c||x^2+xc+c^2|.$$
Let $\delta=\min\{1,\epsilon/x^2+xc+c^2\}$.
Then if $0<|x-c|<\delta$ and therefore since $|x-c|<\epsilon/(x^2+xc+c^2)$,
$$|x^3-c^3|=|x-c||x^2+xc+c^2|<ε.$$
Does this make sense or are the steps done in the right way?
 [/INST] Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-c|<\delta$, then $|x^3-c^3|<\epsilon$.
Now,
$
|x^3-c^3| = |x-c||x^2+cx+c^2|
$
If $|x-c|<1$, then
$||x|-|c|| \leq |x-c|<1$
i.e. $||x|-|c||<1 \implies -1<|x|-|c|<1 \implies |x|<|c|+1$
so that
$$
|x^2+cx+c^2|\leq |x|^2+|c||x|+|c|^2 < (|c|+1)^2+|c|(|c|+1)+|c|^2=3|c|^2+3|c|+1,
$$
and so
$$
|x^3-c^3| = |x-c||x^2+cx+c^2|<(3|c|^2+3|c|+1)|x-c|.
$$
Therefore, for every $\epsilon >0$, there exists a $\delta=\inf\left\{1,\frac{\epsilon}{3|c|^2+3|c|+1}\right\}>0$, such that if $ 0<|x-c|<\delta$ then
$\begin{align} |x^3-c^3|&=|x-c||x^2+cx+c^2|\\
&<\frac{\epsilon}{3|c|^2+3|c|+1}\cdot(3|c|^2+3|c|+1)\\
&=\epsilon \end{align}
$
Thus, by the definition of limit of a function
$$
\lim_{x\to c}x^3=c^3.
$$
</s>