<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] Hint: $\delta$ should not be depending of $x$.
Consider $c > 0$ then
$$ c - 1 < x < c+ 1 \implies |x| < c + 1  $$
Then $$\begin{align}|x^3 - c^3| &= |x - c| |x^2 + xc  +c^2| \\&\leq |x-c|\bigg(|c+1|^2 + |c+1||c| + + |c|^2\bigg) \\&=|x-c|\bigg[|c+1|(|c + 1| + |c|) + |c|^2\bigg]\end{align} $$
Take $$\delta = \min \Bigg\{1, \frac{\epsilon}{|c+1|(|c + 1| + |c|) + |c|^2}\Bigg\}$$
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