To find the limit [tex]\(\lim_ {x \rightarrow \infty} \frac{3 x^2-5}{2 x^3+4 x+3}\)[/tex], let's analyze the given function step-by-step:
1. Understanding the degrees of the terms:
- The numerator [tex]\(3x^2 - 5\)[/tex] is a polynomial of degree 2.
- The denominator [tex]\(2x^3 + 4x + 3\)[/tex] is a polynomial of degree 3.
2. Behavior at infinity:
- When [tex]\(x\)[/tex] approaches infinity ([tex]\(x \rightarrow \infty\)[/tex]), higher-degree terms in both the numerator and the denominator will dominate the behavior of the function.
3. Simplifying the function:
To simplify the behavior as [tex]\(x \rightarrow \infty\)[/tex], we can divide both the numerator and the denominator by the highest power of [tex]\(x\)[/tex] present in the denominator, which is [tex]\(x^3\)[/tex]:
[tex]\[
\frac{3 x^2 - 5}{2 x^3 + 4 x + 3} = \frac{\frac{3 x^2 - 5}{x^3}}{\frac{2 x^3 + 4 x + 3}{x^3}}
\][/tex]
Simplifying each term inside the fractions:
[tex]\[
= \frac{\frac{3 x^2}{x^3} - \frac{5}{x^3}}{\frac{2 x^3}{x^3} + \frac{4 x}{x^3} + \frac{3}{x^3}}
\][/tex]
Which reduces to:
[tex]\[
= \frac{\frac{3}{x} - \frac{5}{x^3}}{2 + \frac{4}{x^2} + \frac{3}{x^3}}
\][/tex]
4. Taking the limit:
As [tex]\(x\)[/tex] approaches infinity, the terms [tex]\(\frac{3}{x}\)[/tex], [tex]\(\frac{5}{x^3}\)[/tex], [tex]\(\frac{4}{x^2}\)[/tex], and [tex]\(\frac{3}{x^3}\)[/tex] all approach 0 because the denominator in each of these fractions grows without bounds.
Therefore, as [tex]\(x \rightarrow \infty\)[/tex], the original expression simplifies to:
[tex]\[
\frac{0 - 0}{2 + 0 + 0} = \frac{0}{2} = 0
\][/tex]
So, the limit is:
[tex]\[
\lim_ {x \rightarrow \infty} \frac{3 x^2 - 5}{2 x^3 + 4 x + 3} = 0
\][/tex]
