To find the limit [tex]\(\lim _{x \rightarrow \infty} \frac{5 x^2+2 x-3}{4 x^3+6 x+5}\)[/tex], we will analyze the behavior of the function as [tex]\(x\)[/tex] approaches infinity.
1. Identify the leading terms:
- The numerator: [tex]\(5 x^2 + 2 x - 3\)[/tex]
- The leading term (highest power of [tex]\(x\)[/tex]) is [tex]\(5x^2\)[/tex].
- The denominator: [tex]\(4 x^3 + 6 x + 5\)[/tex]
- The leading term is [tex]\(4x^3\)[/tex].
2. Compare the degrees of the leading terms:
- The degree of the numerator is 2 (from [tex]\(5x^2\)[/tex]).
- The degree of the denominator is 3 (from [tex]\(4x^3\)[/tex]).
- Since the degree of the denominator is higher than the degree of the numerator, we can expect that as [tex]\(x\)[/tex] approaches infinity, the fraction will tend towards zero.
3. To make it more explicit, we can factor out the highest power of [tex]\(x\)[/tex] from both the numerator and denominator:
- Rewrite the expression as:
[tex]\[
\frac{5 x^2+2 x-3}{4 x^3+6 x+5} = \frac{x^2(5 + \frac{2}{x} - \frac{3}{x^2})}{x^3(4 + \frac{6}{x^2} + \frac{5}{x^3})}
\][/tex]
4. Simplify the expression:
- We can separate the leading term's factor:
[tex]\[
= \frac{x^2}{x^3} \cdot \frac{5 + \frac{2}{x} - \frac{3}{x^2}}{4 + \frac{6}{x^2} + \frac{5}{x^3}}
= \frac{1}{x} \cdot \frac{5 + \frac{2}{x} - \frac{3}{x^2}}{4 + \frac{6}{x^2} + \frac{5}{x^3}}
\][/tex]
5. Take the limit as [tex]\(x\)[/tex] approaches infinity:
- As [tex]\(x\)[/tex] approaches infinity, the terms [tex]\(\frac{2}{x}\)[/tex], [tex]\(\frac{3}{x^2}\)[/tex], [tex]\(\frac{6}{x^2}\)[/tex], and [tex]\(\frac{5}{x^3}\)[/tex] will all approach 0.
[tex]\[
\lim_{x \to \infty} \frac{1}{x} \cdot \frac{5 + \frac{2}{x} - \frac{3}{x^2}}{4 + \frac{6}{x^2} + \frac{5}{x^3}} = \frac{1}{\infty} \cdot \frac{5 + 0 - 0}{4 + 0 + 0}
= 0 \cdot \frac{5}{4} = 0
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{x \to \infty} \frac{5 x^2+2 x-3}{4 x^3+6 x+5} = 0
\][/tex]
