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?

Assume that $\delta < 1/2$, then $1/2<x<3/2$.
$$|x^2+x+1|=x^2+x+1 < 5$$
$$  |x-1||x^2+x+1|< \delta |x^2+x+1|<5\delta =\epsilon$$
You can take it from here.
