Proving that $\lim_{x\to 2}(x^2+x-1)=5$ using an $\epsilon$-$\delta$ argument

Question

I'm just working on the construction portion at the moment for $\lim_{x\to 2}(x^2+x-1)=5$.

Then let $\epsilon > 0$ and $|x^2+x-1-5| < \epsilon \iff |x^2+x-6|<\epsilon$.

Then I write $|x-2||x+3| < \epsilon$.

I have no idea what to say after this point, my professor said to restrict one of the absolute value's but I'm still not sure what he's getting at.

Answer 1

Let $\epsilon>0$. So, we want to show that
$$|x-2||x+3|<\epsilon$$ whenever $0<|x-2|<\delta$ for some $\delta>0$. Your professor said that you need to restrict one of the absolute values. He meant that we need to assume first that $|x-2|<1$. This must be the trick so that we can find a bound for the factor $|x+3|$. Now, we get $-1<x-2<1$, that is, $1<x<3$. With this, we get $4<x+3<6$. This means that $|x+3|<6$ whenever $|x-2|<1$. Then we define $$\delta=\min\big\{1,\frac{\epsilon}{6}\big\}.$$ Then $\delta\leq 1$ and $\delta\leq \frac{\epsilon}{6}$. Hence, if $0<|x-2|<\delta$ then of course $|x-2|<1$ so that $|x+3|<6$ and so $$\begin{align} |(x^2+x-1)-5|&=|x^2+x-6|\\ &=|x-2|\cdot|x+3|\\ &<\delta\cdot6\leq \frac{\epsilon}{6}\cdot 6=\epsilon. \end{align}$$ Done.

Answer 2

Informally, if $x$ is near $2$, you know $x+3 \gt 0$ so you can just remove the absolute value bars and $x+3$ won't be too far from $5$. To be careful about it, you can just say that you will require that $\delta \lt 1$ (or anything reasonably small) which is enough to guarantee that $4 \lt x+3 \lt 6$. Then do your computations that will result in a $\delta$ that depends on the $\epsilon$ you are given. Normally $\epsilon$ will be rather small, which will force $\delta$ to be, and the restriction $\delta \lt 1$ doesn't matter. However, if your normal answer were $\delta = \epsilon$, say, a mean spirited adversary could give you $\epsilon = 2$ and have you fail.

Answer 3

Given $\epsilon>0$, we need $\delta>0$ such that if $|x-2|<\delta$, then $|(x^2+x-1)-5|<\epsilon$. Note that $$ |(x^2+x-1)-5|<\epsilon\Longleftrightarrow |x^2+x-6|<\epsilon\Longleftrightarrow |x-2||x+3|<\epsilon. $$ Suppose we first require that $|x-2|<1$; that is, we are letting $\delta=1$ in this case. Then $$ -1<x-2<1\Longleftrightarrow 4<x+3<6\Longleftrightarrow|x+3|<6. $$ Hence, choose $\delta=\min\{1,\frac{\epsilon}{6}\}$. Then, if $|x-2|<\delta$, we have $|x-2|<\frac{\epsilon}{6}$, and $|x+3|<6$. Thus, $$ |(x^2+x-1)-5|=|x^2+x-6|=|x-2||x+3|<\frac{\epsilon}{6}\cdot6=\epsilon, $$ as desired.

Answer 4

You want to show $$ \forall \epsilon > 0, \exists \delta > 0, |x^2+x-1 -L|<\epsilon : 0<|x-2|<\delta. $$

Hence,

$$ |x^2+x-1 -L| = |x+3||x-2|.$$

Since we know $$ 0 < |x-2| < \delta $$

$$ \Rightarrow 0 < (x-2)^2 < \delta $$ Let $\delta < 1$, without loss of generality.

Then $$ 0 < x^2 -4x +4 < 1$$

$$ \Rightarrow 1<x<3$$ $$ \Rightarrow 4<x+3<6$$ Then, $$ |x^2+x-1 -L| < 6\delta < \epsilon $$

Hence $$ \delta < \frac{\epsilon}{6}$$

Take $\delta:=min\{1,\frac{\epsilon}{6}\}$. Then your statement is true.