Let's analyze the options one by one to determine which function has a limit of 0 as $x \to \infty$:

1. $f(x) = x^2$: As $x \to \infty$, $x^2$ grows without bound, so the limit is $\infty$, not 0.
2. $f(x) = \cos(x)$: The function $\cos(x)$ oscillates between -1 and 1 as $x \to \infty$, so it does not settle to 0.
3. $f(x) = 2^{-x}$: As $x \to \infty$, $2^{-x} = \frac{1}{2^x}$ approaches 0 because the denominator grows exponentially.
4. $f(x) = x$: As $x \to \infty$, $x$ grows without bound, so the limit is $\infty$, not 0.
5. $f(x) = \ln(x)$: As $x \to \infty$, $\ln(x)$ also grows without bound, although more slowly than $x$.

The correct answer is:
$f(x) = 2^{-x}$ since this function approaches 0 as $x \to \infty$.

Would you like a more detailed explanation or have any further questions?

Here are five related questions to further your understanding:

1. What is the general behavior of exponential functions like $a^{-x}$ as $x \to \infty$?
2. How do logarithmic functions behave compared to linear functions as $x \to \infty$?
3. Can a trigonometric function like $\cos(x)$ ever have a limit as $x \to \infty$?
4. What is the difference between functions that grow polynomially and those that grow exponentially?
5. Why does $x^2$ grow faster than $\ln(x)$ as $x \to \infty$?

Tip: When analyzing limits, it's helpful to recognize the general growth rates of common functions like polynomials, exponentials, and logarithms.