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.

We must find a $\delta > 0$ such that $|x-1|<\delta \implies \left|\frac {x^3 - 1}{x-1} - 3\right| < \epsilon$
$\left|\frac {x^3 - 1}{x-1} - 3\right|\\
\left|\frac {x^3 - 3x + 2}{x-1}\right|\\
\left|\frac {(x-1)^2(x+2)}{x-1}\right|$
$\left|\frac {(x-1)^2}{x-1}\right| = |x-1|<\delta$ when $x\ne 1$
Let $\delta \le 1$ then $|x+2| \le 4$
$\left|\frac {(x-1)^2(x+2)}{x-1}\right| < 4\delta$
$\delta = \min(1,\frac {\epsilon}{4})$
