lim _(arrow arrow infty )(6-n^3)/(n^3)+5n^(2+2)=

To find the limit \(\lim_{n \to \infty} \frac{6 \cdot n^3}{n^3 + s^2 + 12}\), we can start by analyzing the expression and simplifying it.<br /><br />First, observe that as \(n\) approaches infinity, the term \(n^3\) in both the numerator and the denominator will dominate over the constant terms \(s^2\) and 12. Therefore, we can factor out \(n^3\) from both the numerator and the denominator:<br /><br />\[<br />\lim_{n \to \infty} \frac{6 \cdot n^3}{n^3 + s^2 + 12} = \lim_{n \to \infty} \frac{6 \cdot n^3}{n^3(1 + \frac{s^2}{n^3} + \frac{12}{n^3})}<br />\]<br /><br />Next, simplify the fraction inside the limit:<br /><br />\[<br />= \lim_{n \to \infty} \frac{6 \cdot n^3}{n^3 \left(1 + \frac{s^2}{n^3} + \frac{12}{n^3}\right)} = \lim_{n \to \infty} \frac{6}{1 + \frac{s^2}{n^3} + \frac{12}{n^3}}<br />\]<br /><br />As \(n\) approaches infinity, the terms \(\frac{s^2}{n^3}\) and \(\frac{12}{n^3}\) approach zero because they are divided by \(n^3\), which grows very large. Thus, the expression simplifies to:<br /><br />\[<br />= \frac{6}{1 + 0 + 0} = \frac{6}{1} = 6<br />\]<br /><br />Therefore, the limit is:<br /><br />\[<br />\boxed{6}<br />\]