To evaluate the limit:

$\lim_{x \to -\infty} \frac{7x^3}{x^3 - 3x^2 + 6}$

Step-by-Step Solution:

1. Look at the highest powers of $x$ in both the numerator and denominator:

   • The highest power of $x$ in the numerator is $7x^3$.
   • The highest power of $x$ in the denominator is $x^3$.

2. Factor out $x^3$ from both the numerator and denominator: $\frac{7x^3}{x^3 - 3x^2 + 6} = \frac{x^3(7)}{x^3(1 - \frac{3}{x} + \frac{6}{x^3})}$

3. Simplify the expression: The $x^3$ terms cancel out: $= \frac{7}{1 - \frac{3}{x} + \frac{6}{x^3}}$

4. Take the limit as $x \to -\infty$: As $x \to -\infty$, both $\frac{3}{x}$ and $\frac{6}{x^3}$ approach 0.

   Therefore, the expression becomes: $\frac{7}{1 - 0 + 0} = \frac{7}{1} = 7$

Conclusion:

$\lim_{x \to -\infty} \frac{7x^3}{x^3 - 3x^2 + 6} = 7$

Would you like further details on this or have any questions? Here are a few related questions for you to explore:

1. What happens to this limit as $x \to +\infty$?
2. How does the behavior of rational functions change as $x$ tends to large values (both positive and negative)?
3. Can you apply this approach to a limit with different powers of $x$?
4. What happens when the highest powers of $x$ in the numerator and denominator are not the same?
5. What would the limit be if the expression had a higher degree term in the numerator, like $7x^4$?

Tip: When dealing with rational functions and limits at infinity, focus on the highest degree terms in the numerator and denominator for simplification.