For a given function $f$ and $\varepsilon = 1$, find $\delta$ such that $0<|x-3|<\delta$ implies $|f(x) - 4| < \varepsilon$

Question

I'm currently learning the precise/formal/$\varepsilon-δ$ definition of a limit.

Let $f(x) = x^3 -9x^2 + 27x -23$

We know: $\displaystyle \lim_{x \to 3} f(x)=4$

For each given value of $\varepsilon > 0$, I'd like to find a corresponding value $δ > 0$ such that:

$$0 < |x-3| < δ \implies |f(x)-4| < \varepsilon$$

Here's my professor's steps for when $\varepsilon = 1:$

1. $|f(x)-4|<1$
2. $-1 < f(x)-4 < 1$
3. $3 < f(x) < 5$
4. $2 < x < 4$
5. $-1 < x-3 < 1$

How did he get to step $4$ and $5$? Is it complete at step $5$ and does $δ = 1$?

I'm very confused, I've been doing research on finding $δ$ when $\varepsilon$ is given. I've found very few examples on this topic, and nothing with $f(x)$ when being compared to as less than $\varepsilon$.

I know how to do proofs but I just can't understand this.

Any help is much appreciated!

Answer 1

The reason is simple enough. On the small interval we're concerned about, where $3 < f(x) < 5$, $f$ is invertible and monotone-increasing. You can see this from the graph:

Therefore, we apply the inverse function and preserve the inequality:

$$f^{-1}(3) < f^{-1}(f(x)) < f^{-1}(5)$$

Of course, $f^{-1}(y)$ is just asking "what $x$ ensures that $f(x) = y$?" You can solve this algebraically or by any other method of your preference, but the graph clearly suggests that

$$f^{-1}(3) = 2 \qquad f^{-1}(5) = 4$$

Definitionally, $f^{-1}(f(x)) = x$, so we now have

$$2 < x < 4$$

The subsequent step is just subtracting $3$. This is because we want $x-3$ to pop up, to match our $\delta$ inequality, and because we get

$$-1 < x-3 < 1 \iff |x-3| < 1$$

i.e. because it's nice enough that we get $\pm 1$ on each side of the inequality, very conducive to the use of the absolute value.

It is from this, then, that we are suggested that $\delta = 1$ works.