Chủ đề này có 1 trả lời

[phuongpro]

$$\lim_{x \to +\infty }x\left ( \dfrac{\pi}{4} - arctan x \right )$$

mod: học latex ở đây nè bạn http://diendantoanho...showtopic=63583

[Crystal]

$$\lim_{x \to +\infty }x\left ( \dfrac{\pi}{4} - arctan x \right )$$

Đặt $$t = \arctan x \Rightarrow x = \tan t;x \to + \infty \Rightarrow t \to \dfrac{\pi }{2}$$
Khi đó: $$\mathop {\lim }\limits_{x \to + \infty } x\left( {\dfrac{\pi }{4} - \arctan x} \right) = \mathop {\lim }\limits_{t \to \dfrac{\pi }{2}} \tan t\left( {\dfrac{\pi }{4} - t} \right) = \mathop {\lim }\limits_{t \to \dfrac{\pi }{2}} \left( {\dfrac{\pi }{4} - t} \right).\mathop {\lim }\limits_{t \to \dfrac{\pi }{2}} \tan t$$
Ta có: $$\mathop {\lim }\limits_{t \to \dfrac{\pi }{2}} \left( {\dfrac{\pi }{4} - t} \right) = \boxed{ - \dfrac{\pi }{4}}$$
$$\mathop {\lim }\limits_{t \to \dfrac{\pi }{2}} \tan t = \mathop {\lim }\limits_{t \to \dfrac{\pi }{2}} \dfrac{{\sin t}}{{\cos t}} = \boxed{+ \infty} $$
Vậy $$\mathop {\lim }\limits_{x \to + \infty } x\left( {\dfrac{\pi }{4} - \arctan x} \right) = \boxed{ - \infty }$$