<s>[INST] Is this proof (scratch work) of $\lim_{x\rightarrow c}x^3=c^3$ correct? Is this scratch work for the proof of $\lim_{x\rightarrow c}x^3=c^3$ correct?
We have that  $|x^3-c^3|=|x-c||x^2+xc+c^2|$.
If $|x-c|<|c|$ then $|x|=|x-c+c|\leq |x-c|+|c|<2|c|$, it follows that $|x^2+xc+c^2|\leq |x|^2+|x||c|+|c|^2< 4|c|^2+2|c|^2+|c|^2=7|c|^2$.
So if $|x-c|<|c|$ and $|x-c|<\frac{\epsilon}{7|c|^2}$ we have that  $|x^3-c^3|=|x-c||x^2+xc+c^2|<\frac{\epsilon 7|c|^2}{7|c|^2}=\epsilon$.
So if we let $\epsilon>0$ be given and $\delta=\min(|c|,\frac{\epsilon}{7|c|^2})$ then $|x^3-c^3|<\epsilon$ if $0<|x-c|<\delta$.
 [/INST] (After reading the comments.) With a little more care, you can include the case $c=0$:
If $|x-c|\le\delta\le1$ then $|x|=|x-c+c|\leq |x-c|+|c|\le1+|c|,$ so
$|x^3-c^3|\le\delta\left(3(1+|c|)^2\right),$
which is $\le\epsilon$ if $\delta$ is not only $\le1$ but also $\le\frac\epsilon{3(1+|c|)^2}.$
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