от stflyfisher » 13 Яну 2011, 12:10
[tex]\lim_{x \to \infty } x(2arctg x-\pi )=[0.\infty ]=\lim_{x \to \infty } \frac {2arctg x-\pi}{\frac{1}{x}}[\frac{0}{0}]=\lim_{x \to \infty }\frac{\frac{2}{1+x^2}-0}{-\frac{1}{x^2}}=-2\lim_{x \to \infty }\frac{x^2}{1+x^2}=[\frac{\infty }{\infty }]=-2\lim_{x \to \infty }\frac{x^2}{x^2(\frac{1}{x^2}+1)}=[/tex]
[tex]=-2\lim_{x \to \infty }\frac{1}{\frac{1}{x^2}+1}=-2.1=-2[/tex]