6. Вычислить: а) limn→∞ \(\frac{-7 n^{4} +6 n^{2} -1}{8 n^{4}-n+6}\) б) limn→∞ \(\frac{2 n^{2}-1}{n^{2}+5}\)

Пошаговое решение:

• а) \( \lim_{n \to \infty} \frac{-7 n^{4} +6 n^{2} -1}{8 n^{4}-n+6} = \lim_{n \to \infty} \frac{\frac{-7 n^{4}}{n^4} + \frac{6 n^{2}}{n^4} -\frac{1}{n^4}}{\frac{8 n^{4}}{n^4}-\frac{n}{n^4}+\frac{6}{n^4}} = \lim_{n \to \infty} \frac{-7 + \frac{6}{n^2} -\frac{1}{n^4}}{8-\frac{1}{n^3}+\frac{6}{n^4}} = \frac{-7 + 0 - 0}{8 - 0 + 0} = -\frac{7}{8} \)
• б) \( \lim_{n \to \infty} \frac{2 n^{2}-1}{n^{2}+5} = \lim_{n \to \infty} \frac{\frac{2 n^{2}}{n^2}-\frac{1}{n^2}}{\frac{n^{2}}{n^2}+\frac{5}{n^2}} = \lim_{n \to \infty} \frac{2-\frac{1}{n^2}}{1+\frac{5}{n^2}} = \frac{2 - 0}{1 + 0} = 2 \)

Ответ: а) \(-\frac{7}{8}\), б) 2.