To find the limit of the function as [tex]\( n \)[/tex] approaches infinity:
[tex]\[
\lim_{n \to \infty} \frac{2n^3 - 2n + n^2}{6n^3}
\][/tex]
we can follow a step-by-step approach.
1. Rewrite the fraction:
We can start by rewriting the given function by separating each term in the numerator:
[tex]\[
\frac{2n^3 - 2n + n^2}{6n^3} = \frac{2n^3}{6n^3} - \frac{2n}{6n^3} + \frac{n^2}{6n^3}
\][/tex]
2. Simplify each term:
Now we simplify each fraction individually:
[tex]\[
\frac{2n^3}{6n^3} = \frac{2}{6} = \frac{1}{3}
\][/tex]
[tex]\[
\frac{2n}{6n^3} = \frac{2}{6n^2} = \frac{1}{3n^2}
\][/tex]
[tex]\[
\frac{n^2}{6n^3} = \frac{1}{6n}
\][/tex]
Thus, we can rewrite our limit as:
[tex]\[
\lim_{n \to \infty} \left( \frac{1}{3} - \frac{1}{3n^2} + \frac{1}{6n} \right)
\][/tex]
3. Analyze the behavior of each term as [tex]\( n \to \infty \)[/tex]:
As [tex]\( n \)[/tex] approaches infinity:
- The term [tex]\(\frac{1}{3}\)[/tex] remains [tex]\(\frac{1}{3}\)[/tex] because it's a constant.
- The term [tex]\(- \frac{1}{3n^2}\)[/tex] approaches 0 because the denominator (which is [tex]\(3n^2\)[/tex]) grows very large.
- The term [tex]\(\frac{1}{6n}\)[/tex] also approaches 0 because the denominator (which is [tex]\(6n\)[/tex]) grows very large.
So, we have:
[tex]\[
\lim_{n \to \infty} \left( \frac{1}{3} - \frac{1}{3n^2} + \frac{1}{6n} \right) = \frac{1}{3} - 0 + 0 = \frac{1}{3}
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{n \to \infty} \frac{2n^3 - 2n + n^2}{6n^3} = \frac{1}{3}
\][/tex]
This concludes our solution. The final answer is [tex]\(\frac{1}{3}\)[/tex].
