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.
 A: 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 $\epsilon>0$.
A: 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 $|x+2| \le 4$
$\left|\frac {(x-1)^2(x+2)}{x-1}\right| < 4\delta$
$\delta = \min(1,\frac {\epsilon}{4})$
A: You want to choose $\delta$ with $|x - 1| \leq \delta$. You also know that $|x + 2| \leq |x - 1| + 3$ (triangle inequality, as Ben commented). So putting that together with what you have you get that you need to be able to choose $\delta(\delta + 3) < \epsilon$, which you can of course do.
A: You don't need the triangle inequality. Starting from your work, assume $\delta < 1$. Then
$$|x-1|<1 \iff 0 < x < 2$$
Therefore $2 < x+2 < 4$, so
$$|(x+2)(x-1)| < 4\cdot \delta\overset{set}{=}\epsilon$$
You want this to be less than $\epsilon$, so choose $\delta$ to be anything less than or equal to $\epsilon/4$. We made the assumption $\delta$ was less than 1, so take it to be the minimum of the 2 options.
Note: same conclusion as Doug M.
