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

To evaluate the limit:<br /><br />\[<br />\lim_{n \to \infty} \frac{6 - n^3}{n^3 + 5n^2 + 2}<br />\]<br /><br />### Step-by-step solution:<br />1. **Factor out \(n^3\) from both the numerator and denominator**:<br /> Divide every term in the numerator and denominator by \(n^3\):<br /><br /> \[<br /> \frac{6 - n^3}{n^3 + 5n^2 + 2} = \frac{\frac{6}{n^3} - 1}{1 + \frac{5}{n} + \frac{2}{n^3}}<br /> \]<br /><br />2. **Simplify as \(n \to \infty\)**:<br /> - As \(n \to \infty\), \(\frac{6}{n^3} \to 0\), \(\frac{5}{n} \to 0\), and \(\frac{2}{n^3} \to 0\).<br /> - The expression simplifies to:<br /><br /> \[<br /> \frac{0 - 1}{1 + 0 + 0} = \frac{-1}{1} = -1<br /> \]<br /><br />### Final Answer:<br />\[<br />\boxed{-1}<br />\]