Задача 74510 ...
Условие
Решение
Все решения
[m]x-2=t[/m] при [m] x → 2[/m]
[m]t → 0[/m]
[m]x=t+2[/m]
[m]lim_{t → 0}(\frac{cos(t+2)}{cos2})^{\frac{1}{t}}=(\frac{cos(0+2)}{cos2})^{\frac{1}{0}}=1^{ ∞ }[/m]
неопределенность.
Применяем второй замечательный предел:
[m]lim_{ α → 0}(1+ α )^{\frac{1}{ α }}=e[/m]
[m]=lim_{t → 0}(1+\frac{cos(t+2)}{cos2}-1)^{\frac{1}{t}}=[/m]
[m] α =1+\frac{cos(t+2)-cos2}{cos2}
[m]=lim_{t → 0}(1+\frac{cos(t+2)-cos2}{cos2})^{\frac{1}{t}}=[/m]
[m]=lim_{t → 0}((1+\frac{cos(t+2)-cos2}{cos2})^{\frac{cos2}{cos(t+2)-cos2}})^{\frac{cos(t+2)-cos2}{cost}\cdot \frac{1}{t}}=[/m]
[m]= lim _{t → 0}((1+\frac{cos(t+2)-cos2}{cos2})^{\frac{cos2}{cos(t+2)-cos2}})^{lim_{t → 0}\frac{cos(t+2)-cos2}{cost}\cdot \frac{1}{t}}=[/m]
[m]=e^{lim_{t → 0}\frac{cos(t+2)-cos2}{cost}\cdot \frac{1}{t}}=[/m] применяем формулу [m]cos α -cos β =[/m]
[m]=e^{lim_{t → 0}\frac{-2sin\frac{t+2-2}{2}sin\frac{t+2+2}{2}}{cost}\cdot \frac{1}{t}}=[/m]
[m]=e^{-2lim_{t → 0}\frac{sin\frac{t}{2}}{\frac{t}{2}}\cdot lim_{t → 0}\frac{ sin\frac{t+4}{2}}{cost}}=[/m]
применяем первый замечательный предел
[m]=lim_{t → 0}\frac{sint}{t}=1[/m]
= [m]=e^{-2\cdot \frac{ sin\frac{0+4}{2}}{cost}}=e^{-\frac{sin2}{cos2}}=e^{-tg2}[/m]