The given limit is:

$\lim_{x \to \infty} \frac{8x^2 - 3}{7x^2 + x - 3}$

Step-by-step solution:

1. Factor the highest degree of $x$ from both the numerator and the denominator. The highest power of $x$ in this case is $x^2$.

   $\frac{8x^2 - 3}{7x^2 + x - 3} = \frac{x^2(8 - \frac{3}{x^2})}{x^2(7 + \frac{1}{x} - \frac{3}{x^2})}$

2. Simplify the expression by canceling out $x^2$ from both the numerator and denominator:

   $= \frac{8 - \frac{3}{x^2}}{7 + \frac{1}{x} - \frac{3}{x^2}}$

3. Evaluate the limit as $x \to \infty$. Notice that as $x$ becomes larger, terms like $\frac{1}{x}$ and $\frac{1}{x^2}$ approach 0.

   • $\frac{3}{x^2} \to 0$
   • $\frac{1}{x} \to 0$

   Thus, the expression simplifies to:

   $\frac{8 - 0}{7 + 0 - 0} = \frac{8}{7}$

Final Answer:

$\lim_{x \to \infty} \frac{8x^2 - 3}{7x^2 + x - 3} = \frac{8}{7}$

Would you like more details or have any questions? Here are some related questions to explore:

1. What happens if the degree of the numerator is lower than the denominator when finding limits at infinity?
2. How does dividing each term by the highest power of $x$ help in simplifying limits at infinity?
3. What are the limits of functions as $x \to -\infty$?
4. Can we apply the same method for rational functions with higher degrees, like $x^3$?
5. How would the limit change if we had a linear term in both the numerator and denominator?

Tip: When evaluating limits at infinity, always check the highest power of $x$ in both the numerator and the denominator—this helps to simplify the expression quickly!