Задача №1049
Условие
Найти предел \(\lim_{n\to\infty}\frac{\sqrt[4]{n^5+2}-\sqrt[3]{n^2+1}}{\sqrt[5]{n^4+2}-\sqrt{n^3+1}}\).
Решение
\[
\lim_{n\to\infty}\frac{\sqrt[4]{n^5+2}-\sqrt[3]{n^2+1}}{\sqrt[5]{n^4+2}-\sqrt{n^3+1}}
=\lim_{n\to\infty}\frac{\frac{\sqrt[4]{n^5+2}}{n^{\frac{3}{2}}}-\frac{\sqrt[3]{n^2+1}}{n^{\frac{3}{2}}}}{\frac{\sqrt[5]{n^4+2}}{n^{\frac{3}{2}}}-\frac{\sqrt{n^3+1}}{n^{\frac{3}{2}}}}=\\
=\lim_{n\to\infty}\frac{\sqrt[4]{\frac{n^5+2}{n^6}}-\sqrt[3]{\frac{n^2+1}{n^{\frac{9}{2}}}}}{\sqrt[5]{\frac{n^4+2}{n^{\frac{15}{2}}}}-\sqrt{\frac{n^3+1}{n^3}}}
=\lim_{n\to\infty}\frac{\sqrt[4]{\frac{1}{n}+\frac{2}{n^6}}-\sqrt[3]{\frac{1}{n^{\frac{5}{2}}}+\frac{1}{n^{\frac{9}{2}}}}}{\sqrt[5]{\frac{1}{n^{\frac{7}{2}}}+\frac{2}{n^{\frac{15}{2}}}}-\sqrt{1+\frac{1}{n^3}}}
=0.
\]
Ответ:
0