What is the Least Upper bound on $\delta$ in the $\epsilon-\delta$ definition of limit? I came upon this question in a GRE Math Subject Test Preparation book (REA).
In the $(\epsilon, \delta)$ definition of limit, $\displaystyle\lim_{x\to c} f(x) = L$, let $f(x) = x^3 + 3x^2 - x + 1$ and let $c = 2$. Find the least upper bound on $\delta$ so that $f(x)$ is bounded within $\epsilon$ of $L$ for all sufficiently small $\epsilon > 0$.
(A) $\frac{\epsilon}{3}$
(B) $\frac{\epsilon}{23}$
(C) $\frac{\epsilon^3}{4}$
(D) $\frac{\epsilon^2}{16}$
(E) $\frac{\epsilon}{19}$
The "explanation" given in the book is:
This is very tedious by direct means, but generally the bound is $\frac{\epsilon}{|f'(c)|}$ when f is differentiable and $f'(c) \neq 0$, on an interval containing c. In this case, $f'(2) = 23$.
Could someone explain why $\frac{\epsilon}{|f'(c)|}$ works, or if there is a better way to do the question?
Update: I've been reading Calculus by Spivak where he finds the limits of polynomials (4th ed, pg. 92-93) and I feel like I have some more intuition.
Let $|x - 2| < u$. By the triangle inequality, $|x| < u + 2$, which implies that $|x^2 + 5x + 9| < (u + 2)^2 + 5 (u + 2) + 9$
Since $f(x) - 19 = (x - 2)(x^2 + 5x + 9)$. If we consider $\delta = \min\left(u,\frac{\epsilon}{(u + 2)^2 + 5 (u + 2) + 9}\right)$, then $|x - 2| < \delta$ implies that $|f(x) - 19| < \epsilon$.
Now if we consider the set $\left\{\min\left(u,\frac{\epsilon}{(u + 2)^2 + 5 (u + 2) + 9}\right) \colon u > 0\right\}$, then the supremum is something like
$\min\left(0,\frac{\epsilon}{(0 + 2)^2 + 5 (0 + 2) + 9}\right) = \min\left(0,\frac{\epsilon}{23}\right)$.
$\lim_{x \to 2} (x^2 + 5x + 9) =\lim_{x \to 2} \frac{f(x) - 19}{x - 2} = f'(2)$
This still feels very imprecise. Is there a more rigorous explanation?
 A: 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.
