<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] Let $\epsilon > 0$ be arbitrary and start off with $\delta_{1} := 1$ so that we have $|x-2| < \delta_{1} = 1$.
Then $x>1$ and $x < 3$ so that $(x-2)^{2} < 1$ and $6x < 18$.
Then $|(x-2)^{2} + 6x|<19$ and set $\delta _{2}$ so that $|x-2| < \delta_{2}$ where $\delta_{2} := \frac{\epsilon}{19}.$
Hence, if $\delta := \min\{\delta_{1},\delta_{2}\}$, we have the result.
The reason why it is necessary to end with a single $\delta$ is that, for this argument to work, we need $|x-2| < 1$ and $|x-2| < \frac{\epsilon}{19}$.
Then to find a fixed value that will work all the way through, choose the smaller one.
</s>