<s>[INST] Cannot find limit using epsilon delta definition Prove
$$\lim\limits_{x\to 2} x^3 = 8$$
using epsilon delta definition.
I try as below.
Let $\varepsilon>0$. We choose $\delta>0$.
Consider that
\begin{align}
\vert x^3-8\vert &= \vert (x-2) (x^2+2x+4)\vert\\
&=\vert (x-2) \vert\vert(x^2+2x+4) \vert \\
&=\vert (x-2) \vert\vert(x-2)^2+6x \vert .
\end{align}
Now I don't know how to continue this answer. I confused with $6x$.
Anyone can help me?

EDIT: I have tried as below as JC12's answer.
Let $\vert x-2\vert <1$, then $\vert x\vert -2< \vert x-2\vert <1$
then we have $$\vert x\vert -2<1 \iff \vert x\vert<3.$$
Now, $$ \vert(x-2)^2+6x \vert < \vert(3-2)^2+6\cdot 3 \vert =19. $$
Thus, \begin{align} \vert x^3-8\vert&= \vert (x-2)
 \vert\vert(x-2)^2+6x \vert < 19 \vert (x-2) \vert. \end{align}
Now choose $\delta=\min(1,\frac{\varepsilon}{19})$. We have
\begin{align} \vert x^3-8\vert < 19 \vert (x-2) \vert< 19
 \frac{\varepsilon}{19} = \varepsilon. \end{align}
So, we can conclude $\lim\limits_{x\to 2} x^3 =8$.

 [/INST] More generally,
if you want to show that
$\lim_{x \to c} x^n=c^n$,
note that
$x^n-c^n
=(x-c)\sum_{k=0}^{n-1} x^kc^{n-1-k}
$.
If
$c-r < x < c+r$
where
$0 < r< \min(1, x)$
then
$\sum_{k=0}^{n-1} x^kc^{n-1-k}
\lt \sum_{k=0}^{n-1} (c+r)^kc^{n-1-k}
\lt \sum_{k=0}^{n-1} (c+r)^k(c+r)^{n-1-k}
=n(c+r)^n
$
so
$\begin{array}\\
|x^n-c^n|
&=|(x-c)\sum_{k=0}^{n-1} x^kc^{n-1-k}|\\
&=|x-c|\,|\sum_{k=0}^{n-1} x^kc^{n-1-k}|\\
&<|x-c|n(c+r)^{n-1}\\
\end{array}
$
so if
$|x-c|
\le \dfrac{\delta}{n(c+r)^{n-1}}
$
then
$|x^n-c^n|
\lt \delta$.
</s>