Given the functions $f(x) = \frac{x-2}{x+2}$ and $g(x) = \frac{1}{x}$, it seems like you want to find the composition of these functions, specifically $f(g(x))$.

The composition $f(g(x))$ means substituting $g(x)$ into $f(x)$:

$f(g(x)) = f\left(\frac{1}{x}\right)$

Substitute $g(x) = \frac{1}{x}$ into $f(x) = \frac{x-2}{x+2}$:

$f\left(\frac{1}{x}\right) = \frac{\left(\frac{1}{x}\right) - 2}{\left(\frac{1}{x}\right) + 2}$

Now, simplify the expression:

$f\left(\frac{1}{x}\right) = \frac{\frac{1 - 2x}{x}}{\frac{1 + 2x}{x}}$

Since the denominators are the same, they cancel out:

$f\left(\frac{1}{x}\right) = \frac{1 - 2x}{1 + 2x}$

So, the composition of the functions is:

$f(g(x)) = \frac{1 - 2x}{1 + 2x}$

Do you have any questions or would you like a more detailed explanation?

---

Here are some related questions you might find useful:

1. How do you find the domain of $f(g(x))$?
2. Can you determine the range of $f(g(x))$?
3. What happens if you try to find $g(f(x))$?
4. How do you differentiate $f(g(x))$?
5. How would you solve for $x$ if $f(g(x)) = 0$?

Tip: When composing functions, always be cautious about the domains of the original functions. The composition is only valid where both functions are defined.