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 $\lim\limits_{x\to 2} x^3 =8$.
Best Answer
You'll find the same approach with many other questions of a similar type, though you first notice that $|x-2|$ is the "part you want" since it is the same form as $0<|x-a|<\delta$. Say we restrict $|x-2|<1$ (this could be any other number but we just choose $1$ for sake of simplicity). Then $|x|-|2|<|x-2|<1$ and thus $|x|<3$. Then $|(x-2)^2+6x|<|(3-2)^2+6\times3|=19$. Thus you can choose your delta as $\min(1,\frac{\varepsilon}{19})$.
I leave you to finish and formalise the proof. As an added bonus, you could also try proving that $\lim_{x\rightarrow a}x^3=a^3, \forall a\in\mathbb{R}$ using a similar method.