Задача №2009
Условие
Найти предел \(\lim_{x\to{0}}\left(\frac{1+\tg{x}}{1+\sin{x}}\right)^{\frac{1}{\sin{x}}}\).
Решение
\[
\lim_{x\to{0}}\left(\frac{1+\tg{x}}{1+\sin{x}}\right)^{\frac{1}{\sin{x}}}
=\left[1^{\infty}\right]
=\lim_{x\to{0}}\left(1+\frac{1+\tg{x}}{1+\sin{x}}-1\right)^{\frac{1}{\sin{x}}}
=\lim_{x\to{0}}\left(1+\frac{\tg{x}-\sin{x}}{1+\sin{x}}\right)^{\frac{1}{\sin{x}}}=\\
=\lim_{x\to{0}}\left(1+\frac{\tg{x}-\sin{x}}{1+\sin{x}}\right)^{\frac{1+\sin{x}}{\tg{x}-\sin{x}}\cdot\frac{\tg{x}-\sin{x}}{\sin{x}\cdot\left(1+\sin{x}\right)}}
=\lim_{x\to{0}}\left(\left(1+\frac{\tg{x}-\sin{x}}{1+\sin{x}}\right)^{\frac{1+\sin{x}}{\tg{x}-\sin{x}}}\right)^{\frac{\frac{1}{\cos{x}}-1}{1+\sin{x}}}
=e^0
=1.
\]
Ответ:
1