To find the limit of the function [tex]\(\lim_{x \rightarrow \infty} \frac{x^3 + 3x^2 + 1}{x + 2}\)[/tex], we can follow these steps:
1. Identify the degrees of the polynomials:
- The numerator [tex]\( f(x) = x^3 + 3x^2 + 1 \)[/tex] is a polynomial of degree 3.
- The denominator [tex]\( g(x) = x + 2 \)[/tex] is a polynomial of degree 1.
2. Compare the degrees:
- Since the degree of the numerator (3) is greater than the degree of the denominator (1), as [tex]\( x \)[/tex] approaches infinity, the numerator's behavior will dominate the function's behavior.
3. Simplify the expression by dividing each term by [tex]\( x^3 \)[/tex], the highest power of [tex]\(x\)[/tex] in the numerator:
[tex]\[
\frac{x^3 + 3x^2 + 1}{x + 2} = \frac{x^3 \left(1 + \frac{3}{x} + \frac{1}{x^3}\right)}{x \left(1 + \frac{2}{x}\right)}
\][/tex]
4. Rewrite the expression:
[tex]\[
\frac{x^3 \left(1 + \frac{3}{x} + \frac{1}{x^3}\right)}{x \left(1 + \frac{2}{x}\right)} = \frac{x^2 \left(1 + \frac{3}{x} + \frac{1}{x^3}\right)}{1 + \frac{2}{x}}
\][/tex]
5. Analyze the limit of each term as [tex]\( x \to \infty \)[/tex]:
[tex]\[
\lim_{x \to \infty} \left(1 + \frac{3}{x} + \frac{1}{x^3}\right) = 1
\][/tex]
[tex]\[
\lim_{x \to \infty} \left(1 + \frac{2}{x}\right) = 1
\][/tex]
[tex]\[
\lim_{x \to \infty} x^2 = \infty
\][/tex]
6. Combine the limits:
[tex]\[
\lim_{x \to \infty} \frac{x^2 \left(1 + \frac{3}{x} + \frac{1}{x^3}\right)}{1 + \frac{2}{x}} = \lim_{x \to \infty} x^2 \cdot \lim_{x \to \infty} \frac{1 + \frac{3}{x} + \frac{1}{x^3}}{1 + \frac{2}{x}} = \infty
\][/tex]
Therefore, the limit of the given function as [tex]\( x \)[/tex] approaches infinity is [tex]\(\infty\)[/tex]. So, we conclude:
[tex]\[
\lim_{x \to \infty} \frac{x^3 + 3x^2 + 1}{x + 2} = \infty
\][/tex]
