Let's break this down step by step.

We are given the expression:

$\lim_{{x \to +\infty}} \left( \frac{-9 - 7}{-8x + 4 \log_3{x}} \right)$

Let's simplify the expression.

1. The numerator is a constant, $-9 - 7 = -16$. So, the expression becomes:

   $\lim_{{x \to +\infty}} \left( \frac{-16}{-8x + 4 \log_3{x}} \right)$

2. Now, let's consider the denominator:

   • As $x \to +\infty$, the term $-8x$ dominates because it grows much faster than the logarithmic term $4 \log_3{x}$.
   • Therefore, the denominator behaves like $-8x$ as $x$ becomes very large.

So, the expression simplifies to:

$\lim_{{x \to +\infty}} \left( \frac{-16}{-8x} \right)$

3. Simplifying this further:

$\lim_{{x \to +\infty}} \frac{-16}{-8x} = \lim_{{x \to +\infty}} \frac{2}{x}$

4. As $x \to +\infty$, $\frac{2}{x} \to 0$.

Thus, the limit of the entire expression is:

$0$

Final answer: $D \, 0$

Would you like more details or have any questions?

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Here are 5 related questions for practice:

1. What is $\lim_{{x \to +\infty}} \frac{1}{x}$?
2. What happens to the expression if the logarithmic term is stronger than the linear term?
3. How would the limit change if the coefficient of $x$ in the denominator was positive?
4. What is $\lim_{{x \to 0}} \frac{1}{x}$?
5. How does the behavior of logarithmic functions compare to polynomial functions at infinity?

Tip: When evaluating limits at infinity, always identify which terms dominate in growth to simplify the expression.