To solve the limit [tex]\(\lim_{x \to \infty} \frac{8x^2 + 7x - 10}{2x^3 - 3x - 4}\)[/tex], follow these steps:
1. Identify the highest degree terms in the numerator and the denominator:
- The numerator is [tex]\(8x^2 + 7x - 10\)[/tex]. The highest degree term in the numerator is [tex]\(8x^2\)[/tex].
- The denominator is [tex]\(2x^3 - 3x - 4\)[/tex]. The highest degree term in the denominator is [tex]\(2x^3\)[/tex].
2. Divide both the numerator and the denominator by [tex]\(x^3\)[/tex], the highest degree term in the denominator:
[tex]\[
\lim_{x \to \infty} \frac{8x^2 + 7x - 10}{2x^3 - 3x - 4} = \lim_{x \to \infty} \frac{\frac{8x^2}{x^3} + \frac{7x}{x^3} - \frac{10}{x^3}}{\frac{2x^3}{x^3} - \frac{3x}{x^3} - \frac{4}{x^3}}
\][/tex]
Simplify each term inside the fraction:
[tex]\[
= \lim_{x \to \infty} \frac{\frac{8}{x} + \frac{7}{x^2} - \frac{10}{x^3}}{2 - \frac{3}{x^2} - \frac{4}{x^3}}
\][/tex]
3. Evaluate the limit term by term as [tex]\(x \to \infty\)[/tex]:
- As [tex]\(x \to \infty\)[/tex], [tex]\(\frac{8}{x} \to 0\)[/tex], [tex]\(\frac{7}{x^2} \to 0\)[/tex], and [tex]\(\frac{10}{x^3} \to 0\)[/tex].
- Similarly, [tex]\(\frac{3}{x^2} \to 0\)[/tex] and [tex]\(\frac{4}{x^3} \to 0\)[/tex].
Therefore:
[tex]\[
\lim_{x \to \infty} \frac{0 + 0 - 0}{2 - 0 - 0} = \frac{0}{2} = 0
\][/tex]
Thus, the limit is:
[tex]\(\lim_{x \to \infty} \frac{8x^2 + 7x - 10}{2x^3 - 3x - 4} = 0\)[/tex]
