To evaluate the limit:

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

we will simplify the expression by factoring out the highest power of $x$ from both the numerator and the denominator.

Step-by-Step Solution:

1. Rewrite the expression: Divide both numerator and denominator by $x^3$ (the highest power of $x$):

   $\frac{7x^3}{x^3 - 3x^2 + 6} = \frac{7}{1 - \frac{3}{x} + \frac{6}{x^3}}.$

2. Analyze as $x \to \infty$:

   • $\frac{3}{x} \to 0$,
   • $\frac{6}{x^3} \to 0$.

   Therefore, the denominator simplifies to:

   $1 - \frac{3}{x} + \frac{6}{x^3} \to 1.$

3. Simplify the fraction: The numerator is $7$, and the denominator approaches $1$, so the entire expression approaches:

   $\frac{7}{1} = 7.$

Final Answer:

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

Let me know if you'd like further explanation or a different approach! 😊

Related Questions:

1. What is the behavior of the same limit if the numerator were $7x^4$?
2. How does factoring help simplify limits as $x \to \infty$?
3. What happens to this limit if the denominator's highest degree term changes?
4. How can you use L'Hôpital's Rule to solve this type of problem?
5. Why do lower-degree terms become negligible as $x \to \infty$?

Tip:

When evaluating limits involving rational functions at infinity, always compare the highest degree terms in the numerator and denominator—they dominate the behavior of the function!