To determine the limits as $x \to \infty$ for the given functions, we'll analyze each one individually.

I. $f(x) = \frac{\ln x}{x^{99}}$

As $x \to \infty$, the numerator $\ln x$ grows slowly compared to the denominator $x^{99}$, which grows extremely fast. So, this expression approaches 0. Specifically:

$\lim_{x \to \infty} \frac{\ln x}{x^{99}} = 0$

Thus, the limit for this function is 0.

II. $f(x) = \frac{e^x}{\ln x}$

As $x \to \infty$, the exponential function $e^x$ grows much faster than $\ln x$. So, this expression tends to infinity:

$\lim_{x \to \infty} \frac{e^x}{\ln x} = \infty$

Thus, the limit for this function is not 0.

III. $f(x) = \frac{x^{99}}{e^x}$

As $x \to \infty$, the exponential function $e^x$ grows significantly faster than any polynomial (like $x^{99}$), causing the fraction to approach 0:

$\lim_{x \to \infty} \frac{x^{99}}{e^x} = 0$

Thus, the limit for this function is 0.

Conclusion

The functions for which the limit is 0 are I and III. Therefore, the correct answer is:

$\boxed{E}$

Do you want further details or have any questions?

---

Here are 5 related questions:

1. What is the limit of $\frac{x^{50}}{e^x}$ as $x \to \infty$?
2. How does the growth of logarithmic functions compare to polynomials?
3. Can $\lim_{x \to \infty} \frac{e^x}{x^k} = 0$ for any constant $k$?
4. Why does $\ln x$ grow slower than $x$?
5. What is the behavior of $\frac{x}{e^x}$ as $x \to \infty$?

Tip: Exponential functions grow faster than any polynomial or logarithmic function as $x \to \infty$.