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I know that $\epsilon$ > 0, and that $\delta$ > 0

I have found this post which deals to a similar practice problem I am having, but the thing is my problem uses 3 roots, and I have no idea what to do after this point.

$$\lim_{x\to7}\left(x^3-9x^2-x+15\right)\ = -90$$

Here you can get three roots, which are |(x-5)(x-7)(x+3)| < $\epsilon$

In general, my main question is how do I know which 2 roots to use and why?

Proving a limit using epsilon delta definition

Asaf Karagila
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3 Answers3

3

$$|(x-5)(x-7)(x+3)|\leq (|x-7|+2)|x-7| (|x-7|+10)$$ So $|x-7| <\delta$ gives $$|(x-5)(x-7)(x+3)| <(\delta+2)\delta (\delta+10)$$ In particular, if $\delta <1$ this gives $$|(x-5)(x-7)(x+3)| <(3)(\delta) (11)$$ so it is enough to take $\delta <1$ and $\delta <\frac {\epsilon} {33}$.

Kavi Rama Murthy
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You need to use all the roots. Here I will use the method outlined in the most upvoted answer (because it is indeed the most straightforward).

First suppose $|x-7| <\delta$. Furthermore we restrict $\delta < 1$, so we only consider $6 < x < 8$.

Using the factorization $|x^3 - 9x^2-x+15+90| = |x-5||x-7|||x+3|$, we still need to bound $|x-5|$ and $|x+3|$. But with the restriction $6 < x < 8$, we easily see that $|x-5| < 3$ and $|x+3| < 11$.

Now we end up with

$$|x^3 - 9x^2-x+15+90| = |x-5||x-7|||x+3| < 3 (\delta)(11) = 33\delta$$

So for any $\epsilon$, we can choose $\delta = \min\{1, \dfrac\epsilon {33}\}$.

player3236
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  • Ah! I see, I have a question with 5 roots later on in my package. We would still do the same steps right?

    And is there ever is a case not to use a |x-c| at any point of solving any of these problems? My tutor asked me that question before our session stopped and it as kept me wondering since.

     –  Oct 09 '20 at 07:57
  • @charming2123 If they are rational functions (polynomial divided by polynomial) you should be able to use $|x-c|$ (but remember the bound for $\delta$ cannot exceed any $c$.) Also remember to reverse the inequality if the factors are in the denominators. – player3236 Oct 09 '20 at 08:00
  • @vb628, right but through the usage of (f(x) - L) < e, you can into these numbers, in order to get the factorization of the new result given. Maybe am I misunderstanding something? –  Oct 09 '20 at 08:03
  • @player3236 I'm just confused about one thing though, why is 2 a min? why not 7? –  Oct 09 '20 at 08:37
  • Oops, I meant $1$ actually, since I restricted $\delta < 1$. – player3236 Oct 09 '20 at 08:38
  • @player3236 still, why 1 in that case? Is it to get the smallest error? –  Oct 09 '20 at 08:39
  • I chose $1$ because it can give us the range of $x$: $6 < x < 8$. This range help us restrict $|x+3|$ and $|x-5|$, which is not directly bounded by $\delta$ without some extra manipulations. – player3236 Oct 09 '20 at 08:41
  • Ohhh I see, I completely forgot about this. Thank you. –  Oct 09 '20 at 08:43
1

You do not need a factorisation. Let $\epsilon>0$. For $|x-7|<\delta\le 1$, we have $|x|<8$, so $$|x^3-7^3| =|x-7||x^2+7x+7^2|\le|x-7|\underbrace{(8^2 + 7\times 8 + 49)}_{=:C_1}\le C_1 \delta,\\ |-9x^2 + 9\times 7^2| = 9|x+7||x-7| \le \underbrace{9 \times (8+7)}_{=:C_2} |x-7| \le C_2 \delta, $$ so if $|x-7|\le \delta:=\frac{\min(\epsilon,1)}{C_1 + C_2 + 1}\le 1$, then as $-90 = 7^3 -9\times 7^2 -7 + 15, $

\begin{align} |x^3 - 9x^2 -x+15 - (-90)| &\le |x^3-7^3| + |-9x+9\times 7^2| + |-x+7| \\ &\le \frac{C_1\min(\epsilon,1)}{C_1+C_2+1} + \frac{C_2\min(\epsilon,1)}{C_1+C_2+1} + \frac{\min(\epsilon,1)}{C_1+C_2+1} \\ &=\min(\epsilon,1)\\ &\le \epsilon. \end{align} This is easier to replicate when factorisation is hard e.g. the degree of the polynomial is large, or you're checking the limit at 'unusual' points. (This is just product rule and sum rule repeatedly applied.)

Calvin Khor
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