Задача с Лопитал

[thepunisher90]

Моля помогнете ми за задачата?
lim x(П/2-arctgx)
x-+∞

[Реклама]

[Добромир Глухаров]

[tex]\lim_{x \to +\infty }\ \ x\ \ (\frac{\pi }{2 } -arctg\ x)=[\infty .0]=\lim_{x \to +\infty }\frac{\frac{\pi }{2 } -arctg\ x}{\frac{1}{x } }=[/tex]
[tex]=[\frac{0}{0 } ]=\lim_{x \to +\infty }\frac{-\frac{1}{1+x^2 }} {-\frac{1}{x^2 } }=\lim_{x \to +\infty }\frac{x^2}{1+x^2}=[/tex]
[tex]=\lim_{x \to +\infty }\frac{1}{\frac{1}{x^2 }+1 }=\frac{1}{(\lim_{x \to +\infty }\frac{1}{x^2 })+1 } =\frac{1}{0+1 } =1[/tex]