Задача 42722 Вычислите пределы...
Условие
Решение
[m]\lim_{x \to \infty }(\frac{2x-1}{2x+1})^{\frac{x+1}{2}}=\lim_{x \to \infty }((1-\frac{2}{2x+1})^{-\frac{2x+1}{2}})^{-\frac{2}{2x+1}\cdot\frac{x+1}{2}= }[/m]
[m]\lim_{x \to \infty }((1-\frac{2}{2x+1})^{-\frac{2x+1}{2}})^{\lim_{x \to \infty }(-\frac{x+1}{2x+1}})=e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}}[/m]
11)
[m]\lim_{x \to 0}\frac{1-e^{-x}}{arctg2x}=\lim_{x \to 0}\frac{1-e^{-x}}{-x}\cdot \frac{2x}{arctg2x}\cdot( -\frac{1}{2})=1\cdot 1\cdot(-\frac{1}{2})=-\frac{1}{2}[/m]
12)
[m]y=(cosx+sinx)^{\frac{1}{x}}[/m]
Логарифмируем:
[m]lny=\frac{1}{x}ln(cosx+sinx)[/m]
[m]\lim_{x \to 0}lny=\lim_{x \to 0}\frac{1}{x}(cosx+sinx)=\lim_{x \to 0}\frac{cosx+sinx}{x}=\frac{1+0}{0}=\infty[/m]
⇒
[m]\lim_{x \to 0}y=\lim_{x \to 0}(cosx+sinx)^{\frac{1}{x}}=e^{\infty }=\infty[/m]