The problem you provided asks to find the limit of the given function as $x \to \infty$:

$f(x) = \frac{3e^{-14x} - e^{18x}}{e^{-x} - 2e^{20x} - e^{-9x}}$

To solve this, we analyze the behavior of the exponential terms in both the numerator and denominator as $x$ approaches infinity.

Step-by-step solution:

1. Examine the terms as $x \to \infty$:

   • In the numerator: $3e^{-14x} \to 0, \quad e^{18x} \to \infty$ So, the numerator is dominated by $-e^{18x}$ as $x \to \infty$.
   • In the denominator: $e^{-x} \to 0, \quad -2e^{20x} \to -\infty, \quad e^{-9x} \to 0$ So, the denominator is dominated by $-2e^{20x}$ as $x \to \infty$.

2. Dominant terms:

   • Numerator: $-e^{18x}$
   • Denominator: $-2e^{20x}$

3. Simplifying the limit: $\lim_{x \to \infty} \frac{-e^{18x}}{-2e^{20x}} = \lim_{x \to \infty} \frac{e^{18x}}{2e^{20x}} = \lim_{x \to \infty} \frac{1}{2e^{2x}} \to 0$

Thus, the limit is:

$\lim_{x \to \infty} f(x) = 0$

Would you like further details or have any questions?

Related Questions:

1. How does the limit change if the exponentials in the denominator are different?
2. What happens if you change the powers of $x$ in the exponentials?
3. Can you apply L'Hopital's rule to this limit?
4. What is the significance of exponential decay in this context?
5. How would the limit behave as $x \to -\infty$?

Tip:

When dealing with limits involving exponentials as $x \to \infty$, focus on the fastest-growing (or fastest-decaying) terms, as they dominate the behavior of the function.