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I've been trying to work through this and got stuck on a step. My professor wrote out a solution but I don't understand it. Would someone help explain what exactly he did?

$|x^2+x-1-5|<\epsilon \impliedby |(x+3)(x-2)|<\epsilon$

Up to this point things make sense to me: we're confirming a limit by searching for a valid $\delta$ (ideally we will find that $|x-2| <$ something involving $\epsilon$). However, my professor suggests the next step to be $|(2+3)^2(x-2)|< \epsilon$, which doesn't make much sense to me. Why did he do this, and what is the reasoning that makes this an appropriate step? Any input is appreciated. Thanks!

Simply Beautiful Art
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Theo C.
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    Basically $\lim_{x\to2} x+3 = 5$ does not strongly influence the choice of $\epsilon$, however doing the same thing for $x - 2$ would result in multiplying by 0. Which does not help to find the $\epsilon$ – Gregory Jun 02 '17 at 01:07
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    $$|(x+3)(x-2)|=|x+3||x-2|<(\delta+5)\delta=\epsilon$$

    when $|x-2|<\delta$. This would've been my approach.

     – Simply Beautiful Art Jun 02 '17 at 01:09
  • @SimplyBeautifulArt: one should avoid the approach you mention. Definition of limit is not supposed to be an exercise in algebraic manipulation of inequalities and in particular one does not solve $\delta$ in terms of $\epsilon$. See this https://math.stackexchange.com/a/2294403/72031 Also your approach will lead to an expression for $\delta$ involving square roots and that is kind of circular. – Paramanand Singh Jun 02 '17 at 08:20
  • @ParamanandSingh Ah, okay. That does make sense, though its just my way of gaining good intuition. If I were really going to do this, I'd probably use it to find a simpler form, like $6\delta=\epsilon\forall\delta<1$. – Simply Beautiful Art Jun 02 '17 at 12:17

3 Answers3

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We will arbitrarily assume that $\delta\leq 1$ (This is a valid assumption to make since, in general, once we find a $\delta$ that works, all smaller values of $\delta$ also work.). Then $|x-2|<\delta\leq 1$ implies that $1< x < 3$ so that $4 < x+3| < 6$ it follows that $$|x+3||x-2| < 6|x-2|<\epsilon$$ iff $|x-2|<\frac{\epsilon}{6}$. Now choose $\delta = \min\{1,\frac{\epsilon}{6}\}$ (This guarantees that both assumptions made about $\delta$ in the course of this proof are taken into account simultaneously.).

Paramanand Singh
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Ahmed
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We compute the limit of $f (x) $ near $a=2$, this means that $x $ is not far away from $2$. So we can assume that the distance from $x $ to $2$ which is $|x-2|$ is for example less than $1$.

thus

$$-1\le x-2\le 1 \tag 1$$ $$\implies 1\le x\le 3$$ $$\implies 4\le x+3 \le 6$$ $$\implies |x+3|\le 6$$ $$\implies |(x+3)(x-2)|\le 6|x-2| \tag 2$$

finally, given $\epsilon>0$,

If $|x-2|\le 1$ and $|x-2|\le \frac {\epsilon}{6} $ then $|f (x)-f (2)|\le \epsilon $.

We can take $$\eta=\min (1,\frac {\epsilon}{6}) .$$ to satisfy conditions $(1) $ and $(2) $.

hamam_Abdallah
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Your Professor might have made the 'suggestion' to get you to realize there can be different ways to handle epsilon/delta arguments, all equally valid. Sometimes there is no 'formula' to plug into.

When working with epsilon/delta arguments, it starts with

$\forall \epsilon, \; \epsilon \gt 0$

You can change this to

$\forall \epsilon, \; 0 \lt \epsilon \le K$

for any (fixed) positive K, with no harm done.

Let $\epsilon$ be any positive number; we can assume that $\epsilon \le 50$

Consider the equation

$|(x+3)| \, |(x−2)| = 50$

Notice that plugging in $7$ for $x$ works.

Explain why choosing $\delta = \frac{\epsilon}{10}$ works for this (almost) arbitrarily given $\epsilon$.

Hint: Determine the maximum value of $|x+3|$ on the interval $[2 - \delta, 2 + \delta]$ by using the fact that $\epsilon \le 50$.

(inequality logic will be provided if requested)

CopyPasteIt
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