By the mean value theorem $|f(x)-L|=|x-c|\cdot f'(\xi)$ for some $\xi$ between $x$ and $c$.
If $\epsilon, \delta$ are small then $f'(\xi)\approx f'(c)$ for all $\xi\in (c-\delta,c+\delta)$ and hence $|x-c|<\frac \epsilon{|f'(c)|}$ implies $|f(x)-L|\stackrel <\approx \epsilon $.
Or to put it differently: Near $c$ the function $f(x)$ approximately looks like $g(x)=f(c)+(x-c)f'(c)$ and for $g$ it is clear that $\delta=\frac \epsilon{|f'(c)|}$ is the best choice.
Let's make the above rigorous:
We are given an open interval $I\subseteq \mathbb R$ and a continuoiusly differentiable function $f\colon I\to \mathbb R$ and a point $c\in I$ and with $f'(c)\ne 0$.
Consider the set
$$\mathcal D = \{h\colon (0,\infty)\to(0,\infty)\mid\forall\epsilon>0\colon\forall x\in I\colon |x-c|<h(\epsilon)\Rightarrow |f(x)-f(c)|<\epsilon\}.$$
Then $\mathcal D$ contains all functions that one is allowed to use for finding a suitable $\delta$, given $\epsilon$ in determining $\lim_{x\to c}f(x)$.
Unfortunately, there is no natural order relation on $\mathcal D$, hence it is not immediatel clear how to apply terms like "least upper bound" to $\mathcal D$.
However, we have a partial order (not only on $\mathcal D$, but on the set of all functions $(0,\infty)\to(0,\infty)$) as we can define $h_1\le h_2$ if $\limsup_{x\to 0}\frac{h_1(x)}{h_2(x)}\le 1$.
Let $H(x)=\frac x{|f'(c)|}$ and $\tilde h(\epsilon)=\inf\{|x-c|\mid |f(x)-f(c)|\ge\epsilon\}$.
Note that in general $H\notin \mathcal D$.
We prove the following statements:
- $\tilde h\in\mathcal D$
- If $h\in \mathcal D$ then $H\ge h$
- $\tilde h\ge H$
These three statements together allow us to loosely speak of $H(\epsilon)$ being an approximation to the maximal possible choice of $\delta$ for sufficiently small $\epsilon$.
Lemma 1: $\tilde h\in\mathcal D$.
Proof: Since $f$ is continuous, for $\epsilon>0$ there exists $\delta>0$ such that $|x-c|<\delta$ implies $|f(x)-f(c)|<\epsilon$. We conclude that $\tilde h(\epsilon)\ge \delta>0$, hence $h$ is indeed a function $(0,\infty)\to (0,\infty)$.
If $\epsilon>0$ is given and $|x-c|<\tilde h(\epsilon)$, then $|f(x)-f(c)|<\epsilon$, as otherwise we would have $\tilde h(\epsilon)\le |x-c|$. Therefore $\tilde h \in \mathcal D$.$_\blacksquare$
Lemma 2: If $h\in \mathcal D$ then $H\ge h$.
Proof:
We have to show that for $\epsilon>0$ there exists $\delta>0$ such that $0<x<\delta$ implies $\frac{H(x)}{h(x)}>1-\epsilon$, i.e. $H(x)>h(x)(1-\epsilon)$.
I am a bit too tired for these calculations right now, so I will add them tomorrow - one has to be careful not to mix $\epsilon, \delta$ belonging to the $\limsup$ with those belonging to $h$ or the continuity of $f$.
Lemma 3: $\tilde h\ge H$.
Proof:
Same here, see you tomorrow.
Take the case where $x\in [-2,0]$ (i.e. $\delta < 1$). Then we have the following inequality
$$\left|\frac{(x+1)(2x-1)}{3(x^2-x+1)}\right| < \frac{5}{3}|x+1|$$
by maximizing the numerator and minimizing the denominator. So set $\delta = \min(\frac{3}{5}\epsilon,1)$ and the proof step for the limit follows in both cases.
If $\epsilon > \frac{5}{3}$:
$$|x+1|<1\implies \left|\frac{(x+1)(2x-1)}{3(x^2-x+1)}\right| < \frac{5}{3}|x+1| < \frac{5}{3} < \epsilon$$
If $\epsilon \leq \frac{5}{3}$:
$$|x+1|<\frac{3}{5}\epsilon \implies \left|\frac{(x+1)(2x-1)}{3(x^2-x+1)}\right| < \frac{5}{3}|x+1| < \epsilon$$
Best Answer
You're right to want to bound values of $x$ away from $-1$, and to do so by $\frac{1}{2}$ is fine. The inequality you obtain by requiring $x$ to be within $\frac{1}{2}$ of $2$ is
$$-\frac{5}{2}<x<-\frac{3}{2}$$
which you have written. Then, as you've written, we have
$$-\frac{3}{2}<x+1<-\frac{1}{2}$$
Notice that the above inequality implies that $|x+1|>\frac{1}{2}.$ It follows that
$$\frac{5|x+2|}{|x+1|}<10|x+2|$$
So let $\delta(\epsilon)<\min\{\frac{1}{2},\frac{\epsilon}{10}\}$.