Given certain conditions for $\delta$, how do I show that an inequality relating delta to x is true?

This is a problem out of a textbook (though there's no answer to this one in the back).

If   $0 < \delta < 1$ and $|x-4| < \delta$

show:   $|\sqrt{x}-2| < \frac{\delta}{\sqrt{3} + 2}$

Hint: Rationalize $|\sqrt{x} - 2|$.

I rationalize $|\sqrt{x} - 2|$ by multiplying by a form of 1, $\frac{\sqrt{x} + 2}{\sqrt{x} + 2}$.

This yields $|\frac{x-4}{\sqrt{x} + 2}| < \frac{\delta}{\sqrt{3} + 2}$. (1)

Given the conditions from the beginning of the problem, I think that:

$0 < \delta < 1 , |x - 4| < \delta$   can be rewritten as   $-\delta < x - 4 < \delta$...

...which can then be rewritten as $4 - \delta< x < 4 + \delta$.

Given this and (1):

$$4-\left|\frac{(x)-4}{\sqrt{x} - 2}\right| (\sqrt{3} + 2) < x < 4 + \left|\frac{(x)-4}{\sqrt{x} - 2}\right|(\sqrt{3} + 2)$$

Further, because I know that $0 < \delta < 1 , |x - 4| < \delta$ means that $|x - 4| < 1$ must be true.

The absolute value inequality can be rewritten as   $-1 < x - 4 < 1$...

...or $3 < x < 5$.

By plugging possible values for x in, I get: $$4-\left|\frac{(3)-4}{\sqrt{3} - 2}\right| (\sqrt{3} + 2) < x < 4 + \left|\frac{(3)-4}{\sqrt{3} - 2}\right|(\sqrt{3} + 2)$$ and $$4-\left|\frac{(5)-4}{\sqrt{5} - 2}\right| (\sqrt{3} + 2) < x < 4 + \left|\frac{(5)-4}{\sqrt{5} - 2}\right|(\sqrt{3} + 2)$$

Simplifying, I find that:

$$4 + \left|\frac{\sqrt{3} + 2}{\sqrt{3}-2}\right| < x < 4 - \left|\frac{\sqrt{3} + 2}{\sqrt{3}-2}\right|$$

and

$$4 - \left|\frac{\sqrt{5} + 2}{\sqrt{5}-2}\right| < x < 4 + \left|\frac{\sqrt{5} + 2}{\sqrt{5}-2}\right|$$

What I'd like to know is: am I even barking up the right tree?

If so, how might I proceed?

If not, where did I go astray (and how do I go about understanding this problem and other problems like it?)

I feel like I don't really understand the question being asked, and I'm just manipulating the values I've been given because busy work feels like progress.

I will admit that (even after a semester of calculus) I'm still at a bit of a loss as to what a $\epsilon-\delta$ proof (or the "precise definition of the limit [...]", which I can only do because I've practiced a few problems while working from reference material) is useful for (and what it means beyond mapping some arbitrary values to some other arbitrary values).

I appreciate the time y'all have spent reading this - and especially any help you can offer.

You look as if you're trying to start with $|\sqrt{x}-2| < \frac{\delta}{\sqrt{3} + 2}$ and end up with $|x-4|\lt\delta$. But, the problem is asking for it the other way around. That is, we must start with: $$|x-4|\lt\delta$$ Thus (assuming $x\ge0$, which is valid because $x$ is in $4\pm1$): $$\left|(\sqrt{x}-2)(\sqrt{x}+2)\right|\lt\delta$$
As $\sqrt{x}+2\ge0$ for all $x\ge0$: $$\left|(\sqrt{x}-2)\right|(\sqrt{x}+2)\lt\delta$$
Dividing over: $$\left|(\sqrt{x}-2)\right|\lt\frac{\delta}{\sqrt{x}+2}$$
Since $x$ is within $4\pm1$, we have: $$\frac{\delta}{\sqrt{x}+2}\le\frac{\delta}{\sqrt{4-1}+2}$$
This implies: $$\left|\sqrt{x}-2\right|\lt\frac{\delta}{\sqrt{3}+2}$$