Finding supremum of all $\delta > 0$ for the $(\epsilon , \delta)$-definition of $\lim_{x \to 2} x^3 + 3x^2 -x + 1$

In the $(\epsilon , \delta)$-definition of the limit, $$\lim_{x \to c} f(x) = L,$$ 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 $\epsilon > 0$.

I know the definition and here it’s clear that $L = 23$. But when I tried to calculate $\delta$ directly from the definition it was too complex for me to solve.

According to one solution it said that usually for a differentiable $f(x)$, $\delta = \epsilon \ |f'(c)|$.
I have never seen this fact before and since I don’t know what theorem it is, it would be great if I someone could direct me to where I can learn about this fact.

It would even be better if one could show me how to do it directly, as well.

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You’re right that the direct $\epsilon$-$\delta$ can be bothersome.

Recall the

Mean Value Theorem: Let $f$ be continous on $[a,b]$ and differentiable on $(a,b)$. Then

$$f(b)-f(a)=f'(c)(b-a)$$

for some $c\in(a,b)$.

In particular, if your $f$ is continuously differentiable on $\Bbb R$, as is the case of polynomials, for any fixed interval $[x,y]$ you will be able to find what $M=\sup_{t\in[x,y]}f(t)$ is, so $$|f(w)-f(z)|\leq M|w-z|$$ for any $w,z\in[x,y]$. Thus, if we make $|w-z|<\epsilon /M$; we’re done, see?

Note that, for every $|x|\leqslant3$,
$$
|f(x)-f(2)|=|x-2|\cdot|x^2+5x+9|\leqslant|x-2|\cdot(3^2+5\cdot3+9)=33\cdot|x-2|.
$$
Hence, for each $\epsilon\gt0$, a suitable choice of $\delta$ is
$$
\delta=\min\{1;\epsilon/33\}.
$$

From basic analysis we know that, $\forall x$, $\exists \epsilon > 0$ such that $|f(x)-L| < \epsilon$ whenever $ |x -c | < \delta$ for some $\delta>0$.
Now, $|f(x) – L| = |x^3+3x^2−x+1-(8+12-2+1)| = |x^3+3x^2-x-16|$
It is obvious that 2 is a root for the above function, using this fact and the triangle inequality we have
$|x^3+3x^2-x-16| = |x-2|*|x^2+5x+9| \leq |4+20+9||x-2| = 33|x-2|$ since $|x-2| <\delta$