giới hạn lớp 11 (2)

Question

1) Tìm các giới hạn sau:

$a/ \mathop {\lim }\limits_{x \to \frac{\pi }{2}}\frac{1-sinx}{(\frac{\pi }{2}-x)^2}; b/\mathop {\lim }\limits_{x \to \frac{\pi }{6}}\frac{sin(x-\frac{\pi }{6})}{\frac{\sqrt{3}}{2}-cosx}.$

score 2 · Answer 1

Bình chọn tăng 2 Bình chọn giảm

Câu b.

Đặt $u_{n}=\frac{sin(x-\frac{\pi}{6})}{\frac{\sqrt{3}}{2}-cosx}$

n \to \infty

Ta đặt $2t=x-\frac{\pi}{6}=>x=2t+\frac{\pi}{6}$ (Khi $x \to \frac{\pi}{6} \Leftrightarrow t \to 0)$

$u_{n} =\frac{sin2t}{cos\frac{\pi}{6}-cos(2t+\frac{\pi}{6})}=\frac{sin2t}{2sin(\frac{\pi}{6}+t)sin2t}=\frac{1}{2sin(\frac{\pi}{6}+t)}$

Ta có

$\mathop {\lim }\limits_{t \to 0}u_{n}=\mathop {\lim }\limits_{x \to \frac{\pi}{6}}\frac{(\frac{\pi}{6}+t)}{2(\frac{\pi}{6}+t).sin(\frac{\pi}{6}+t)}=\frac{3}{\pi}$

lịch sử

Nếu thấy đúng thì vote up nhé! Nếu thấy sai thì nhắc nhở, ko sửa thì vote down thẳng tay :)) – Nero 28-03-14 04:24 PM

score 3 · Answer 2

Bình chọn tăng 3 Bình chọn giảm

Câu a.

$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{1-sinx}{(\frac{\pi}{2}-x)^2}$

=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{(cos\frac{x}{2}-sin\frac{x}{2})^2}{(\frac{\pi}{2}-x)^2}$

=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{2sin^2[\frac{1}{2}(\frac{\pi}{2}-x)]}{(\frac{\pi}{2}-x)^2}$

=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{\frac{1}{2}(\frac{\pi}{2}-x)^2.[\frac{sin[\frac{1}{2}(\frac{\pi}{2}-x)}{\frac{1}{2}(\frac{\pi}{2}-x)}]^2}{(\frac{\pi}{2}-x)^2}$

=$\frac{1}{2}$

lịch sử