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?