Alright, let's solve the limit problem step by step:
We need to find the limit
[tex]\[
\lim _{n \rightarrow+\infty} \frac{2 n^2 + n - 1}{n^3 - 6 n^2 + n + 2}
\][/tex]
Step-by-step solution:
1. Identify the degrees of the polynomial in the numerator and the denominator:
The highest degree term in the numerator [tex]\(2n^2 + n - 1\)[/tex] is [tex]\(2n^2\)[/tex].
The highest degree term in the denominator [tex]\(n^3 - 6n^2 + n + 2\)[/tex] is [tex]\(n^3\)[/tex].
2. Divide both the numerator and the denominator by the highest degree term in the denominator (which is [tex]\(n^3\)[/tex]):
[tex]\[
\lim_{n \to \infty} \frac{2n^2 + n - 1}{n^3 - 6n^2 + n + 2} = \lim_{n \to \infty} \frac{\frac{2n^2}{n^3} + \frac{n}{n^3} - \frac{1}{n^3}}{\frac{n^3}{n^3} - \frac{6n^2}{n^3} + \frac{n}{n^3} + \frac{2}{n^3}}
\][/tex]
3. Simplify the fractions in both the numerator and the denominator:
[tex]\[
= \lim_{n \to \infty} \frac{\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3}}{1 - \frac{6}{n} + \frac{1}{n^2} + \frac{2}{n^3}}
\][/tex]
4. Analyze the behavior of each term as [tex]\(n\)[/tex] approaches infinity:
- [tex]\(\frac{2}{n} \rightarrow 0\)[/tex]
- [tex]\(\frac{1}{n^2} \rightarrow 0\)[/tex]
- [tex]\(\frac{1}{n^3} \rightarrow 0\)[/tex]
- [tex]\(\frac{6}{n} \rightarrow 0\)[/tex]
- [tex]\(\frac{1}{n^2} \rightarrow 0\)[/tex]
- [tex]\(\frac{2}{n^3} \rightarrow 0\)[/tex]
5. Substitute these limits back into the expression:
[tex]\[
= \frac{0 + 0 - 0}{1 - 0 + 0 + 0} = \frac{0}{1} = 0
\][/tex]
Therefore, the limit is:
[tex]\[
\lim _{n \rightarrow+\infty} \frac{2 n^2 + n - 1}{n^3 - 6 n^2 + n + 2} = 0
\][/tex]
