Задача №1048
Условие
Найти предел \(\lim_{n\to\infty}\frac{\sqrt{n^3-2n^2+1}+\sqrt[3]{n^4+1}}{\sqrt[4]{n^6+6n^5+2}-\sqrt[5]{n^7+3n^3+1}}\).
Решение
\[
\lim_{n\to\infty}\frac{\sqrt{n^3-2n^2+1}+\sqrt[3]{n^4+1}}{\sqrt[4]{n^6+6n^5+2}-\sqrt[5]{n^7+3n^3+1}}
=\lim_{n\to\infty}\frac{\frac{\sqrt{n^3-2n^2+1}+\sqrt[3]{n^4+1}}{n^{\frac{3}{2}}}}{\frac{\sqrt[4]{n^6+6n^5+2}-\sqrt[5]{n^7+3n^3+1}}{n^{\frac{3}{2}}}}=\\
=\lim_{n\to\infty}\frac{\sqrt{\frac{n^3-2n^2+1}{n^3}}+\sqrt[3]{\frac{n^4+1}{n^{\frac{9}{2}}}}}{\sqrt[4]{\frac{n^6+6n^5+2}{n^6}}-\sqrt[5]{\frac{n^7+3n^3+1}{n^{\frac{15}{2}}}}}
=\lim_{n\to\infty}\frac{\sqrt{1-\frac{2}{n}+\frac{1}{n^3}}+\sqrt[3]{\frac{1}{n^{\frac{1}{2}}}+\frac{1}{n^{\frac{9}{2}}}}}{\sqrt[4]{1+\frac{6}{n}+\frac{2}{n^6}}-\sqrt[5]{\frac{1}{n^{\frac{1}{2}}}+\frac{3}{n^{\frac{9}{2}}}+\frac{1}{n^{\frac{15}{2}}}}}
=1.
\]
Ответ:
1