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To find the limit [tex]\(\lim _{x \rightarrow \infty} \frac{x^3+3 x^2+1}{x+2}\)[/tex], we can follow a detailed and step-by-step approach:
1. Understand the Function Behavior as [tex]\(x \to \infty\)[/tex]:
The given function is a rational function. Rational functions are defined as the ratio of two polynomials. Here, we have the polynomial in the numerator [tex]\( P(x) = x^3 + 3x^2 + 1 \)[/tex] and the polynomial in the denominator [tex]\( Q(x) = x + 2 \)[/tex].
2. Dominant Terms:
As [tex]\( x \)[/tex] approaches infinity, we observe the highest degree terms in both the numerator and the denominator since they will dominate the behavior of the function. In our numerator, the highest degree term is [tex]\( x^3 \)[/tex] and in the denominator, it is [tex]\( x \)[/tex].
3. Simplify by Factoring Out Dominant Terms:
To simplify:
[tex]\[ \frac{x^3 + 3x^2 + 1}{x + 2} \][/tex]
we factor out [tex]\( x^3 \)[/tex] from the numerator and [tex]\( x \)[/tex] from the denominator:
[tex]\[ = \frac{x^3(1 + \frac{3x^2}{x^3} + \frac{1}{x^3})}{x(1 + \frac{2}{x})} = \frac{x^3(1 + \frac{3}{x} + \frac{1}{x^3})}{x(1 + \frac{2}{x})} \][/tex]
4. Simplify the Fraction:
[tex]\[ = \frac{x^3(1 + \frac{3}{x} + \frac{1}{x^3})}{x(1 + \frac{2}{x})} = \frac{x^2(1 + \frac{3}{x} + \frac{1}{x^3})}{1 + \frac{2}{x}} \][/tex]
5. Taking the Limit:
As [tex]\( x \)[/tex] approaches infinity, the terms [tex]\(\frac{3}{x}\)[/tex], [tex]\(\frac{1}{x^3}\)[/tex], and [tex]\(\frac{2}{x}\)[/tex] approach 0. Therefore, the function simplifies to:
[tex]\[ = \frac{x^2(1 + 0 + 0)}{1 + 0} = x^2 \][/tex]
6. Conclusion:
The limit of [tex]\( x^2 \)[/tex] as [tex]\( x \)[/tex] approaches infinity is obviously infinity. Thus, we conclude:
[tex]\[ \lim _{x \rightarrow \infty} \frac{x^3+3 x^2+1}{x+2} = \infty \][/tex]
So the limit, [tex]\(\lim _{x \rightarrow \infty} \frac{x^3+3 x^2+1}{x+2}\)[/tex], is [tex]\(\infty\)[/tex].
1. Understand the Function Behavior as [tex]\(x \to \infty\)[/tex]:
The given function is a rational function. Rational functions are defined as the ratio of two polynomials. Here, we have the polynomial in the numerator [tex]\( P(x) = x^3 + 3x^2 + 1 \)[/tex] and the polynomial in the denominator [tex]\( Q(x) = x + 2 \)[/tex].
2. Dominant Terms:
As [tex]\( x \)[/tex] approaches infinity, we observe the highest degree terms in both the numerator and the denominator since they will dominate the behavior of the function. In our numerator, the highest degree term is [tex]\( x^3 \)[/tex] and in the denominator, it is [tex]\( x \)[/tex].
3. Simplify by Factoring Out Dominant Terms:
To simplify:
[tex]\[ \frac{x^3 + 3x^2 + 1}{x + 2} \][/tex]
we factor out [tex]\( x^3 \)[/tex] from the numerator and [tex]\( x \)[/tex] from the denominator:
[tex]\[ = \frac{x^3(1 + \frac{3x^2}{x^3} + \frac{1}{x^3})}{x(1 + \frac{2}{x})} = \frac{x^3(1 + \frac{3}{x} + \frac{1}{x^3})}{x(1 + \frac{2}{x})} \][/tex]
4. Simplify the Fraction:
[tex]\[ = \frac{x^3(1 + \frac{3}{x} + \frac{1}{x^3})}{x(1 + \frac{2}{x})} = \frac{x^2(1 + \frac{3}{x} + \frac{1}{x^3})}{1 + \frac{2}{x}} \][/tex]
5. Taking the Limit:
As [tex]\( x \)[/tex] approaches infinity, the terms [tex]\(\frac{3}{x}\)[/tex], [tex]\(\frac{1}{x^3}\)[/tex], and [tex]\(\frac{2}{x}\)[/tex] approach 0. Therefore, the function simplifies to:
[tex]\[ = \frac{x^2(1 + 0 + 0)}{1 + 0} = x^2 \][/tex]
6. Conclusion:
The limit of [tex]\( x^2 \)[/tex] as [tex]\( x \)[/tex] approaches infinity is obviously infinity. Thus, we conclude:
[tex]\[ \lim _{x \rightarrow \infty} \frac{x^3+3 x^2+1}{x+2} = \infty \][/tex]
So the limit, [tex]\(\lim _{x \rightarrow \infty} \frac{x^3+3 x^2+1}{x+2}\)[/tex], is [tex]\(\infty\)[/tex].
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