To find the limit [tex]\(\lim _{n \rightarrow+\infty} \frac{2 n^2+n-2}{n^3-n^2+n+2}\)[/tex], we can approach the problem by following a step-by-step method.
1. Identify dominant terms: When [tex]\( n \)[/tex] approaches infinity, the dominant terms in both the numerator and the denominator will govern the behavior of the expression. Here, the dominant term in the numerator is [tex]\(2n^2\)[/tex], and the dominant term in the denominator is [tex]\(n^3\)[/tex].
2. Simplify the expression: Divide both the numerator and the denominator by the highest power of [tex]\( n \)[/tex] present in the denominator, which is [tex]\( n^3 \)[/tex].
[tex]\[
\frac{2n^2 + n - 2}{n^3 - n^2 + n + 2} \div \frac{1}{n^3} = \frac{\frac{2n^2}{n^3} + \frac{n}{n^3} - \frac{2}{n^3}}{\frac{n^3}{n^3} - \frac{n^2}{n^3} + \frac{n}{n^3} + \frac{2}{n^3}}
\][/tex]
3. Rewrite the expression:
[tex]\[
= \frac{\frac{2}{n} + \frac{1}{n^2} - \frac{2}{n^3}}{1 - \frac{1}{n} + \frac{1}{n^2} + \frac{2}{n^3}}
\][/tex]
4. Evaluate the limit of each individual term: As [tex]\( n \)[/tex] approaches infinity, each term containing [tex]\( \frac{1}{n} \)[/tex], [tex]\( \frac{1}{n^2} \)[/tex], and higher powers of [tex]\( n \)[/tex] will approach zero.
[tex]\[
\lim_{n \to \infty} \left(\frac{\frac{2}{n} + \frac{1}{n^2} - \frac{2}{n^3}}{1 - \frac{1}{n} + \frac{1}{n^2} + \frac{2}{n^3}}\right) = \frac{0 + 0 - 0}{1 - 0 + 0 + 0}
\][/tex]
5. Simplify the final limit:
[tex]\[
= \frac{0}{1} = 0
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{n \to \infty} \frac{2n^2 + n - 2}{n^3 - n^2 + n + 2} = 0
\][/tex]
