<s>[INST] How to use the epsilon delta definition to prove that $\lim\limits_{x\to 1} \frac{x^3-1}{x-1} = 3$
Not sure if I am doing this right, however, this is what I have:
Let $\epsilon > 0$. We need to find a $\delta > 0$ such that $0<|x-1|<\delta$ leads to the conclusion $|f(x)-3|<\epsilon$.
$$|\frac{x^3-1}{x-1} - 3| < \epsilon$$
We know that $x^3-1 = (x-1)(x^2+x+1)$, so
$$|x^2+x+1-3|<\epsilon$$
$$|(x+2)(x-1)|<\epsilon$$
Do I need to use the triangle inequality? Not really sure where to go from here. Any help is appreciated.
 [/INST] Following @Ben's comment...
We want to find $\delta$ such that $|x-1|<\delta \Rightarrow |(x-2)(x-1)|<\epsilon$.
As $|(x-2)(x-1)|=|x-2||x-1|\leq(|x-1|+3)|x-1|$, because of the triangle inequality..then it is enough to find $\delta$ such that
$|x-1|<\delta \Rightarrow (|x-1|+3)|x-1|<\epsilon$
Notice that $(|x-1|+3)|x-1|<\epsilon \iff z^2+3z-\epsilon<0$ when $z=|x-1|\geq0$. From here one can conclude $|x-1|=z \in \Big[0,\frac{-3+\sqrt{9+4\epsilon}}{2}\Big)$ by solving the inequation.
This is equivalent to affirm that $|x-1|<\frac{-3+\sqrt{9+4\epsilon}}{2}$ and so for any $\epsilon>0$ one can take $\delta=\frac{-3+\sqrt{9+4\epsilon}}{2}$. Note that $\frac{-3+\sqrt{9+4\epsilon}}{2}>0$ for any $\epsilon>0$.
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