giới hạn lớp 11 (1)

Question

1) Tìm các giới hạn sau:

$a/ \mathop {\lim }\limits_{x \to \frac{\pi }{4}}\frac{cosx-sin x}{cos2x}; b/ \mathop {\lim }\limits_{x \to 0}\frac{1+sinx-cosx}{1-sinx-cosx}$

score 1 · Answer 1

Bình chọn tăng 1 Bình chọn giảm

a. $A=-\dfrac{\sin x -\cos x}{\cos 2x}=-\dfrac{\sqrt 2 \sin (x-\dfrac{\pi}{4})}{\cos 2x}$ đặt $x-\dfrac{\pi}{4}=t$

$A=-\dfrac{\sqrt 2 \sin t}{\cos (2x-\dfrac{\pi}{2})}=-\dfrac{\sqrt 2 \sin t}{\sin 2t}=-\dfrac{\sqrt 2 }{2\cos t}$

Vậy $\lim_{t\to 0}A=-\dfrac{\sqrt 2}{2}$

score 1 · Answer 2

Bình chọn tăng 1 Bình chọn giảm

$\dfrac{1+\sin x -\cos x}{1-\sin x -\cos x}=\dfrac{2\sin \dfrac{x}{2} \cos \dfrac{x}{2} +2\sin^2 \dfrac{x}{2}}{2\sin^2 \dfrac{x}{2} -2\sin \dfrac{x}{2} \cos \dfrac{x}{2}}=\dfrac{ \cos \dfrac{x}{2} +\sin \dfrac{x}{2}}{\sin \dfrac{x}{2} - \cos \dfrac{x}{2}}$

$\lim_{x\to 0} \dfrac{ \cos \dfrac{x}{2} +\sin \dfrac{x}{2}}{\sin \dfrac{x}{2} - \cos \dfrac{x}{2}}=-1$