Prove the statement using the $\epsilon$, $\delta$ definition of a limit: $$\lim \limits_{x \to 3}{(x^2+x-4)} = 8$$

The Precise Definition of a Limit

In case you're not familiar with the definition of "The Precise Definition of a Limit", here it is.

Let $f$ be a function defined on some open interval that contains the number $a$, except possible $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write $$\lim \limits_{x \to a}{f(x)} = L$$ if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that $$\text{if } 0 < |x - a| < \delta \text{ then } |f(x) - L| < \epsilon$$


$\lim \limits_{x \to 3}{(x^2+x-4)} = 8 \implies \text{if } 0 < |x - 3| < \delta \text{ then } |x^2+x-4 - 8| < \epsilon$

$|x^2+x-4 - 8| < \epsilon \implies |x^2+x-12| < \epsilon \implies |(x-3)(x+4)| < \epsilon$

From here, I quickly get lost.

I notice that $|(x-3)(x+4)| < \epsilon \implies |x-3||x+4| < \epsilon \implies |x-3| < \frac{\epsilon}{|x+4|}$, which means that $\delta = \frac{\epsilon}{|x+4|}$.

In the other exercises, I often end the proof here, because I've solved for $\delta$ purely in terms of $\epsilon$. However, here I have $x$ on the RHS.

  • $\begingroup$ Intuitively, note that x lies close to 3. Assuming delta is smaller than 1 for instance, we know that x lies between 2 and 4, which gives absolute upper and lower bounds on $|x+4|$. This should allow you to express delta only in terms of epsilon. $\endgroup$ – TMM Nov 23 '16 at 2:53
  • $\begingroup$ Hint: you can write it as $(x-3)^2 + 7(x-3) + 8$ which makes the $\epsilon,\delta$ calculations easier. $\endgroup$ – dxiv Nov 23 '16 at 3:31
  • $\begingroup$ That wasn't meant as a - minus sign. I changed it to a , comma. $\endgroup$ – dxiv Nov 23 '16 at 3:35
  • $\begingroup$ Your function $x^2+x-4=(x-3)^2+7(x-3)+8$. $\endgroup$ – dxiv Nov 23 '16 at 3:37
  • $\begingroup$ @dxiv -- Alright, thanks. If you want to post your hint as an answer, I'll go ahead and give it a check. $\endgroup$ – Fine Man Nov 23 '16 at 3:38

Hint: write $f(x)=x^2+x-4=(x-3)^2+7(x-3)+8$ and let $\epsilon \gt 0 $. Then:

$$|f(x)-8| \lt \epsilon \;\;\iff\;\; |(x-3)^2 + 7(x-3)| \lt \epsilon$$

Choose $\delta = \min(1, \epsilon / 8)$ and let $0 \lt |x-3| \lt \delta$. Then:

  • $0 \lt |x-3| \lt \delta \le 1$ implies that $|x-3|^2 \lt |x-3|$

  • $|x-3| \lt \delta \le \epsilon / 8$ implies that $8|x-3| \lt \epsilon$

Using the above and the triangle inequality:

$$|(x-3)^2 + 7(x-3)| \le |x-3|^2 + 7|x-3| \lt |x-3| + 7|x-3| = 8 |x-3| \lt \epsilon$$

  • $\begingroup$ Where'd you get $\delta=\min(1,\epsilon/8)$ from? $\endgroup$ – Fine Man Nov 23 '16 at 5:52
  • $\begingroup$ @SirJony By inspection, lacking a better word. Look at the inequality that needs to hold: $|(x-3)^2 + 7(x-3)| \lt \epsilon$. Then think of how small does $|x-3|$ need to be in order to make sure that the respective inequality holds. Or, work backwards and don't choose $\delta$ except for assuming it to be $\delta \le 1$. Then work out the rest and end up with $|(x-3)^2 + 7(x-3)| \lt 8 \delta$. At that point, choose $\delta$ such that $8 \delta \le \epsilon$ so that the target inequality is satisfied. $\endgroup$ – dxiv Nov 23 '16 at 5:57
  • $\begingroup$ Yes, but how'd you derive the specific numbers? I tried using the quadratic formula on $a^2+7a-\epsilon$ but (obviously) got results other than $1$ and $\epsilon/8$. $\endgroup$ – Fine Man Nov 23 '16 at 6:01
  • $\begingroup$ @SirJony You are (over)complicating it. You don't need exact bounds, just sufficient ones. The above does just that. It amounts to $|a^2 + 7a| \le |a^2| + 7|a| \lt 8|a|$ for $|a| \lt 1$.. $\endgroup$ – dxiv Nov 23 '16 at 6:03
  • 1
    $\begingroup$ Ahh! I get it. Thanks. Let me finish up reading the rest of your answer and I'll give it a checkmark. $\endgroup$ – Fine Man Nov 23 '16 at 6:10

Since you're looking at the limit as $x\to3$ one could assume that as $x$ approaches this value, it is within a distance of say $1$ from $3$. That is, $|x-3|<1$ for sufficiently close $x$ (which is the case since $x\to3$). This inequality gives: $$|x-3|<1\\2<x<4\\6<x+4<8$$ Thus $|x+4|$ is bounded between $6$ and $8$. Returning to what you had: $|x-3|<\frac{\epsilon}{|x+4|}$, we see that the minimum of $\frac{\epsilon}{|x+4|}$ occurs when $|x+4|$ is maximized, which is $8$ (as we showed). Thus$|x-3||x+4|<\epsilon$ so $|x-3|<\frac{\epsilon}{8}$. Now we have two inequalities: $$|x-3|<1$$ and $$|x-3|<\frac{\epsilon}{8}$$ So try choosing $\delta$ as the minimum of $1$ and $\frac{\epsilon}{8}$, and verify that it works.

  • $\begingroup$ $\frac{\epsilon}{|x+4|}<\frac{\epsilon}{6}$ in this case $\endgroup$ – Andrei Nov 23 '16 at 3:07
  • $\begingroup$ Ah yes. I got my minimums and maximums mixed up. Will edit, thank you. $\endgroup$ – Dave Nov 23 '16 at 3:15
  • $\begingroup$ That's not true. Given $\epsilon>0$ and as you suggest choose $\delta=\min\{1,\epsilon/6\}$ then as you showed $6<x+4<8$ and thus: $$|x^2+x-12|<|x-3||x+4|<\frac{\epsilon}{6} 8=4\epsilon/3 \not < \epsilon $$ $\endgroup$ – Ruzayqat Nov 23 '16 at 3:22
  • $\begingroup$ This seems to contradict (or maybe not, correct me if I'm wrong) the definition of "The Precise Definition of a Limit". $\epsilon$ must be able to represent any number, but you restricted it when your presumed $|x−3|<1$. $\endgroup$ – Fine Man Nov 23 '16 at 3:28
  • 1
    $\begingroup$ @SirJony This is a good point. The assumption that $|x-3|<1$ is one that is made to simplify the calculations, but we take care of that assumption at the end. We choose $\delta$ as the minimum of $1$ and the other value dependent on epsilon for that reason. If for some reason we consider a huge $\epsilon$, then $\delta=1$ will certainly work. However when we consider the limit as $x\to 3$, we really only care about small $\epsilon$ values, as we are in a very close neighbourhood around $3$. $\endgroup$ – Dave Nov 23 '16 at 3:53

Given $\epsilon>0$, choose $\delta=\min\{1,\epsilon/8\}$, then $-1<x-3<1$ implies that $2<x<4$. Thus, $6<x+4=|x+4|<8$ and hence to have $|x^2+x-12|<\epsilon$, you need to choose $\delta$ such that $|x-3|<\epsilon/8$.


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