The problem asks to compute the limit:

$\lim_{x \to -\infty} \frac{x + 4}{3x^2 - 5}$

Step-by-Step Solution:

1. Identify the highest powers of $x$ in the numerator and the denominator:

   • The numerator $x + 4$ has the highest power of $x$ as $x^1$.
   • The denominator $3x^2 - 5$ has the highest power of $x$ as $x^2$.

2. Divide both the numerator and the denominator by the highest power of $x$ in the denominator, which is $x^2$: $\frac{x + 4}{3x^2 - 5} = \frac{\frac{x + 4}{x^2}}{\frac{3x^2 - 5}{x^2}} = \frac{\frac{x}{x^2} + \frac{4}{x^2}}{\frac{3x^2}{x^2} - \frac{5}{x^2}}$

   Simplifying this expression: $= \frac{\frac{1}{x} + \frac{4}{x^2}}{3 - \frac{5}{x^2}}$

3. Take the limit as $x \to -\infty$:

   • As $x \to -\infty$, $\frac{1}{x} \to 0$, and $\frac{4}{x^2} \to 0$, and $\frac{5}{x^2} \to 0$.
   • So, the expression becomes: $\lim_{x \to -\infty} \frac{\frac{1}{x} + \frac{4}{x^2}}{3 - \frac{5}{x^2}} = \frac{0 + 0}{3 - 0} = 0$

Final Answer:

$\lim_{x \to -\infty} \frac{x + 4}{3x^2 - 5} = 0$

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

Related Questions:

1. What is the general rule for determining the limit at infinity when the degrees of the numerator and denominator differ?
2. How does dividing by the highest power of $x$ help in solving limits involving infinity?
3. How do limits behave differently when approaching positive versus negative infinity?
4. What happens when the degrees of the numerator and denominator are the same in a rational function limit?
5. How does L'Hopital's Rule apply to limits involving infinity?

Tip:

When evaluating limits at infinity, always focus on the highest degree terms in the numerator and denominator—they dominate the behavior of the function at large values of $x$.