Prove by epsilon-delta that $\lim \limits_{x \to 1} x^3-4=-3$ I need to prove that $\lim \limits_{x \to 1} x^3-4=-3$ with epsilon-delta.
My work
$\forall \varepsilon > 0 ,\exists \space \delta > 0: 0<|x-1|< \delta \implies |x^3-4+3| < \varepsilon$
Working with the consequent:
$|x^3-4+3| <  \varepsilon \iff |x^3-1| < \varepsilon \iff |x-1||x^2+x+1| < \varepsilon$
Multiplying the antecedent by $|x^2+x+1|:$
$|x-1|< \delta$ $/\cdot |x^2+x+1| \iff  |x-1||x^2+x+1|< \delta |x^2+x+1|$
Here i found a relation but i don't know how to proceed, and i don't know if this way is the best way to prove this. Any hints?

Hint: For $\lim_{x\to a}f(x)=L$, we essentially want to show that $|f(x) - L|<\epsilon$ for every $|x-a|<\delta$; that is, we want to show that $|x^3-1|<\epsilon$ for every $|x-1|<\delta$. Let $\delta=1$. Then $|x-1|<\delta$ or simply $|x-1|<1\implies x\in(0,2)$; hence, we also see that $|x^2+x+1|<|2^2+2+1|=7$. This implies that we should choose $\delta=\min\{1,\frac{\epsilon}{7}\}$. Why? Because, if so, then we have
$$
\begin{align*}
|(x^3-4)-(-3)|
&= |x^3-1|\\[0.5em]
&= |x-1||x^2+x+1|\\[0.5em]
&< \frac{\epsilon}{7}\cdot 7\\[0.5em]
&= \epsilon.
\end{align*}
$$
Hopefully that supplies the missing detail you needed.
