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To evaluate the limit:

\[
\lim_{x \to \infty} \frac{5x^3 + 6x^2}{6x^3 - 7}
\]

we can simplify by dividing both the numerator and denominator by \( x^3 \), the highest power of \( x \) present in the expression:

\[
\lim_{x \to \infty} \frac{5 + \frac{6}{x}}{6 - \frac{7}{x^3}}
\]

As \( x \to \infty \), terms with \( \frac{6}{x} \) and \( \frac{7}{x^3} \) approach zero, leaving:

\[
\frac{5}{6}
\]

Thus, the evaluated limit is:

\[
\lim_{x \to \infty} \frac{5x^3 + 6x^2}{6x^3 - 7} = \frac{5}{6}
\]

Would you like more details or have any questions?

### Related Questions:
1. What would change if the limit was \( x \to -\infty \)?
2. How do we handle limits at infinity for expressions with different leading terms?
3. What if the powers of \( x \) in the numerator and denominator were different?
4. How does the limit change if we add a constant to the numerator or denominator?
5. How does this concept apply to finding horizontal asymptotes?

**Tip:** When evaluating limits at infinity, divide by the highest power of \( x \) in the denominator to simplify the expression.

Evaluate the limit \( \lim_{x \to \infty} \frac{5x^3 + 6x^2}{6x^3 - 7} \)

Explore how to evaluate the limit of \( \lim_{x \to \infty} \frac{5x^3 + 6x^2}{6x^3 - 7} \) by dividing by the highest power of x. This example demonstrates techniques for limits at infinity, suitable for calculus students in Grades 11-12.

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