Задача №1119
Условие
Найти предел \(\lim_{x\to{0}}\frac{(1-\cos{x})^2}{\tg^3{x}-\sin^3{x}}\).
Решение
\[
\lim_{x\to{0}}\frac{(1-\cos{x})^2}{\tg^3{x}-\sin^3{x}}
=\left|\frac{0}{0}\right|
=\lim_{x\to{0}}\frac{(1-\cos{x})^2\cdot\cos^3{x}}{\sin^3{x}\cdot\left(1-\cos^3{x}\right)}=\\
=\lim_{x\to{0}}\frac{(1-\cos{x})\cdot\cos^3{x}}{\sin^3{x}\cdot\left(1+\cos{x}+\cos^2{x}\right)}
=\lim_{x\to{0}}\frac{(1-\cos{x})\cdot(1+\cos{x})\cdot\cos^3{x}}{\sin^3{x}\cdot\left(1+\cos{x}+\cos^2{x}\right)\cdot(1+\cos{x})}=\\
=\lim_{x\to{0}}\frac{(1-\cos^2{x})\cdot\cos^3{x}}{\sin^3{x}\cdot\left(1+\cos{x}+\cos^2{x}\right)\cdot(1+\cos{x})}
=\lim_{x\to{0}}\frac{\cos^3{x}}{\sin{x}\cdot\left(1+\cos{x}+\cos^2{x}\right)\cdot(1+\cos{x})}
=\infty.
\]
Ответ:
\(\infty\)