$\lim_{x \to 2} x^2+x-1 = 5$ with epsilon-delta

Question

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!

Answer 1

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) $.

Answer 2

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.).

Answer 3

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.

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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)