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Sagot :
Certainly! Let's evaluate the limit [tex]\(\lim _{x \rightarrow \infty} \frac{8 x^2+7 x-10}{2 x^3-3 x-4}\)[/tex] step-by-step.
### Step-by-Step Solution:
1. Identify the degrees of the polynomials in the numerator and the denominator:
- The numerator [tex]\(8x^2 + 7x - 10\)[/tex] is a polynomial of degree 2 (the highest power of [tex]\(x\)[/tex] is [tex]\(x^2\)[/tex]).
- The denominator [tex]\(2x^3 - 3x - 4\)[/tex] is a polynomial of degree 3 (the highest power of [tex]\(x\)[/tex] is [tex]\(x^3\)[/tex]).
2. Compare the degrees:
- Since the degree of the polynomial in the numerator (2) is less than the degree of the polynomial in the denominator (3), this implies that as [tex]\(x\)[/tex] approaches infinity, the numerator grows more slowly than the denominator.
3. Understanding the behavior at infinity:
- When the degree of the denominator is greater than the degree of the numerator, the fraction [tex]\(\frac{8x^2 + 7x - 10}{2x^3 - 3x - 4}\)[/tex] tends toward 0.
4. Formal approach: Divide by the highest power of [tex]\(x\)[/tex] in the denominator:
- Let's divide both the numerator and the denominator by [tex]\(x^3\)[/tex], the highest power of [tex]\(x\)[/tex] in the denominator:
[tex]\[ \lim _{x \rightarrow \infty} \frac{ \frac{8x^2+7x-10}{x^3} }{ \frac{2x^3-3x-4}{x^3} } = \lim _{x \rightarrow \infty} \frac{ \frac{8x^2}{x^3} + \frac{7x}{x^3} + \frac{-10}{x^3} }{ \frac{2x^3}{x^3} + \frac{-3x}{x^3} + \frac{-4}{x^3} } = \lim _{x \rightarrow \infty} \frac{ \frac{8}{x} + \frac{7}{x^2} + \frac{-10}{x^3} }{ 2 + \frac{-3}{x^2} + \frac{-4}{x^3} } \][/tex]
5. Evaluate the limit:
- As [tex]\(x\)[/tex] approaches infinity, [tex]\(\frac{8}{x}\)[/tex], [tex]\(\frac{7}{x^2}\)[/tex], [tex]\(\frac{-10}{x^3}\)[/tex], [tex]\(\frac{-3}{x^2}\)[/tex], and [tex]\(\frac{-4}{x^3}\)[/tex] all approach 0.
- Thus the expression simplifies to:
[tex]\[ \lim _{x \rightarrow \infty} \frac{ \frac{8}{x} + \frac{7}{x^2} + \frac{-10}{x^3} }{ 2 + \frac{-3}{x^2} + \frac{-4}{x^3} } = \frac{0 + 0 + 0}{2 + 0 + 0} = \frac{0}{2} = 0 \][/tex]
### Conclusion:
Therefore, the limit is:
[tex]\[ \boxed{0} \][/tex]
### Step-by-Step Solution:
1. Identify the degrees of the polynomials in the numerator and the denominator:
- The numerator [tex]\(8x^2 + 7x - 10\)[/tex] is a polynomial of degree 2 (the highest power of [tex]\(x\)[/tex] is [tex]\(x^2\)[/tex]).
- The denominator [tex]\(2x^3 - 3x - 4\)[/tex] is a polynomial of degree 3 (the highest power of [tex]\(x\)[/tex] is [tex]\(x^3\)[/tex]).
2. Compare the degrees:
- Since the degree of the polynomial in the numerator (2) is less than the degree of the polynomial in the denominator (3), this implies that as [tex]\(x\)[/tex] approaches infinity, the numerator grows more slowly than the denominator.
3. Understanding the behavior at infinity:
- When the degree of the denominator is greater than the degree of the numerator, the fraction [tex]\(\frac{8x^2 + 7x - 10}{2x^3 - 3x - 4}\)[/tex] tends toward 0.
4. Formal approach: Divide by the highest power of [tex]\(x\)[/tex] in the denominator:
- Let's divide both the numerator and the denominator by [tex]\(x^3\)[/tex], the highest power of [tex]\(x\)[/tex] in the denominator:
[tex]\[ \lim _{x \rightarrow \infty} \frac{ \frac{8x^2+7x-10}{x^3} }{ \frac{2x^3-3x-4}{x^3} } = \lim _{x \rightarrow \infty} \frac{ \frac{8x^2}{x^3} + \frac{7x}{x^3} + \frac{-10}{x^3} }{ \frac{2x^3}{x^3} + \frac{-3x}{x^3} + \frac{-4}{x^3} } = \lim _{x \rightarrow \infty} \frac{ \frac{8}{x} + \frac{7}{x^2} + \frac{-10}{x^3} }{ 2 + \frac{-3}{x^2} + \frac{-4}{x^3} } \][/tex]
5. Evaluate the limit:
- As [tex]\(x\)[/tex] approaches infinity, [tex]\(\frac{8}{x}\)[/tex], [tex]\(\frac{7}{x^2}\)[/tex], [tex]\(\frac{-10}{x^3}\)[/tex], [tex]\(\frac{-3}{x^2}\)[/tex], and [tex]\(\frac{-4}{x^3}\)[/tex] all approach 0.
- Thus the expression simplifies to:
[tex]\[ \lim _{x \rightarrow \infty} \frac{ \frac{8}{x} + \frac{7}{x^2} + \frac{-10}{x^3} }{ 2 + \frac{-3}{x^2} + \frac{-4}{x^3} } = \frac{0 + 0 + 0}{2 + 0 + 0} = \frac{0}{2} = 0 \][/tex]
### Conclusion:
Therefore, the limit is:
[tex]\[ \boxed{0} \][/tex]
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