To find the limit of the function [tex]\(\frac{x^3 + 3x^2 + 1}{x + 2}\)[/tex] as [tex]\(x\)[/tex] approaches infinity, follow these detailed steps:
1. Divide by the Highest Power of [tex]\(x\)[/tex] in the Denominator:
The numerator is [tex]\(x^3 + 3x^2 + 1\)[/tex], and the denominator is [tex]\(x + 2\)[/tex]. The highest power of [tex]\(x\)[/tex] in the denominator is [tex]\(x\)[/tex].
To simplify, divide both the numerator and the denominator by [tex]\(x\)[/tex]:
[tex]\[
\frac{x^3 + 3x^2 + 1}{x + 2} = \frac{\frac{x^3 + 3x^2 + 1}{x}}{\frac{x + 2}{x}}
\][/tex]
2. Simplify Each Term:
Simplify each term inside the numerator and the denominator separately:
- For the numerator: [tex]\(\frac{x^3 + 3x^2 + 1}{x} = x^2 + 3x + \frac{1}{x}\)[/tex]
- For the denominator: [tex]\(\frac{x + 2}{x} = 1 + \frac{2}{x}\)[/tex]
So the limit expression now becomes:
[tex]\[
\frac{x^2 + 3x + \frac{1}{x}}{1 + \frac{2}{x}}
\][/tex]
3. Examine the Behavior as [tex]\(x \to \infty\)[/tex]:
As [tex]\(x\)[/tex] approaches infinity:
- [tex]\(x^2\)[/tex] grows without bound.
- [tex]\(3x\)[/tex] also grows without bound but slower than [tex]\(x^2\)[/tex].
- [tex]\(\frac{1}{x}\)[/tex] approaches 0.
- [tex]\(\frac{2}{x}\)[/tex] also approaches 0.
Hence, the simplified expression can be considered as:
[tex]\[
\frac{x^2 + 3x + \frac{1}{x}}{1 + \frac{2}{x}} \approx \frac{x^2 + 3x}{1}
\][/tex]
4. Divide the Terms by the Leading Term in the Numerator:
To better understand the limit at infinity, focus on the leading terms:
[tex]\[
\frac{x^2 + 3x}{1} = x^2 \left(1 + \frac{3}{x}\right)
\][/tex]
As [tex]\(x\)[/tex] goes to infinity, [tex]\(\frac{3}{x}\)[/tex] approaches 0, so the expression becomes approximately:
[tex]\[
\frac{x^2 (1 + 0)}{1} = x^2
\][/tex]
5. Recognize the Dominant Term:
Since [tex]\(x^2\)[/tex] is the dominant term in this fraction and grows without bound as [tex]\(x\)[/tex] approaches infinity, the whole expression similarly approaches infinity.
Therefore, the limit is:
[tex]\[
\boxed{\infty}
\][/tex]
