<s>[INST] Prove that $\ f(x)=x^{3}$ is continuous at $\ x = -2$ using the delta-epsilon approach Prove that $\ f(x)=x^{3}$ is continuous at $\ x = -2$.
I am really struggling on how to choose an initial $\delta_1$ value, are we able to choose any value or are there specific values to target, in this case I chose $\delta_1=1$
The below is the work I had: 
For any given $\epsilon > 0$, we want to find a $\delta >0$ such that $\ |f(x) - (-8)| < \epsilon$ whenever $\ 0<|x-(-2)|<\delta$ each simplifies to $\ |x^{3} +8| < \epsilon$ and  $\ 0<|x+2|<\delta$
I expand the inequality so that I get
$$\ |(x+2)(x^{2}-2x+4)|<\epsilon$$
I temporarily set $\ |(x^{2}-2x+4)|<K$ so then the above inequality becomes:
$$\ K|x+2|<\epsilon\\
|x+2|<\frac{\epsilon}{K}$$
This is the part I am struggling with, now I need to find out K.
We know that $\ 0<|x+2|<\delta$, removing the absolute values results in $-\delta<x+2<\delta$ and getting to only x results in $-\delta-2<x<\delta-2$
If I make the assumption of $\delta_1=1$ then the resulting inequality is $-3<x<-1$ 
I also tried this with a $\delta=\frac{1}{2}$ just to make sure it fell inside these bounds.
Since we want to look at the upper bound here I plugged in -1 back into K and got the following
$$\ 
|x+2|<\frac{\epsilon}{|-1|^2+2|-1|+4|}\\
|x+2|<\frac{\epsilon}{7} \\
\delta=\frac{\epsilon}{7}\\
\delta=min(1,\frac{\epsilon}{7})
$$
Then I did the following to prove it
$$
|f(x) - (-8)| =|x^{3}+8|=|(x^{2}-2x+4)||x+2|\\
<|x+2|\cdot (|-1|^{2}+2|-1|+4)\\= |x+2|\cdot7\\
=7\delta \le 7 \cdot \frac{\epsilon}{7}=\epsilon
$$
 [/INST] Note that\begin{align}x^2-2x+4&=\bigl((x+2)-2\bigr)^2-2\bigl((x+2)-2\bigr)+4\\&=(x+2)^2-6(x+2)+12\end{align}and that therefore$$|x+2|<1\implies|x^2-2x+4|<1+6+12=19.$$So, take $\delta=\min\left\{1,\frac\varepsilon{19}\right\}$ and then$$|x+2|<\delta\implies\bigl|(x+2)(x^2-2x+4)\bigr|<\frac\varepsilon{19}\times19=\varepsilon.$$
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