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$\lim\limits_{x\to 1} \frac{x^3-1}{x-1} = 3$

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.

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    $\begingroup$ 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 $. $\endgroup$ Oct 21, 2021 at 22:27

4 Answers 4

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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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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 $|x+2| \le 4$

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

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

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

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

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

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