# 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.

• If you bound $|x-1|$ then $|x+2| \leq |x-1| + 3$. You need to find some expression in $\epsilon$ to bound $|x-1|$ so that then $|x+2||x-1| < \epsilon$. Oct 21, 2021 at 22:27

## 4 Answers

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 $$\epsilon>0$$.

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 $$|x+2| \le 4$$

$$\left|\frac {(x-1)^2(x+2)}{x-1}\right| < 4\delta$$

$$\delta = \min(1,\frac {\epsilon}{4})$$

• Little bit confused how you went from step 1 to 2 to get $x^3 -3x + 2$. Why did you do that as well?
– ST4S
Oct 21, 2021 at 22:52
• What do you do when you are adding rational functions? You find a common denominator. Oct 21, 2021 at 22:57

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.

• So $\delta = \frac{\epsilon}{\delta+3}$? Still not clear to me...
– ST4S
Oct 21, 2021 at 22:56

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.

• That's what I figured out and got the same result! Thanks though :)
– ST4S
Oct 23, 2021 at 20:28