2 replies to this topic

[nguyentmlinh1995]

a. $\lim_{x\rightarrow \frac{\Pi }{6}}\frac{\sqrt{3}- 2cosx}{36x^{2}-\Pi ^{2}}$

b. $\lim_{x\rightarrow 0}\frac{\ln \left ( 1-2\tan ^{2}x \right )}{x^{2}}$

c. $\lim_{x\rightarrow -\frac{\Pi }{4}}\frac{1+\sin 2x}{\cos 2x.\left ( \cos x-\sin x \right )}$

[Gioi han]

a. $\lim_{x\rightarrow \frac{\Pi }{6}}\frac{\sqrt{3}- 2cosx}{36x^{2}-\Pi ^{2}}$

b. $\lim_{x\rightarrow 0}\frac{\ln \left ( 1-2\tan ^{2}x \right )}{x^{2}}$

c. $\lim_{x\rightarrow -\frac{\Pi }{4}}\frac{1+\sin 2x}{\cos 2x.\left ( \cos x-\sin x \right )}$

a. Áp dụng quy tắc $L' Hospital$ ta có:

$\lim_{x\rightarrow \frac{\pi}{6}}\frac{\sqrt{3}-2\cos x}{36x^2-\pi^2}$

$=\lim_{x\rightarrow \frac{\pi}{6}}\frac{2\sin x}{72x}=\frac{1}{12\pi}$

 

b. $\lim_{x\rightarrow 0}\frac{\ln \left( 1- 2\tan ^2x \right)}{-2\tan ^2x}. \frac{-2\tan ^2x}{x^2}= 1. (-2)=-2$

 

c.$ \lim_{x\rightarrow \frac{\pi}{4}}\frac{1+\sin 2x}{\cos 2x( \cos x- \sin x)}$

$=\lim_{x\rightarrow \frac{\pi}{4}}\frac{(\sin x+ \cos x)^2}{(1-2\sin 2x)( \cos x+ \sin x)}$

$=\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos x- \sin x}{1+ 2\cos 2x}( L' Hospital)$

$=\sqrt{2}$

[nguyentmlinh1995]

c.$ \lim_{x\rightarrow \frac{\pi}{4}}\frac{1+\sin 2x}{\cos 2x( \cos x- \sin x)}$

$=\lim_{x\rightarrow \frac{\pi}{4}}\frac{(\sin x+ \cos x)^2}{(1-2\sin 2x)( \cos x+ \sin x)}$

$=\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos x- \sin x}{1+ 2\cos 2x}( L' Hospital)$

$=\sqrt{2}$

Tại sao $\left\{ ( 1 - 2\sin 2x \right \}'$ lại bằng ${\left ( 1-2\cos 2x \right )}'$ mà không phải là $4\cos 2x$ v bạn??

 

@@: Không biết sửa lại đúng như thắc bạn muốn hỏi ko?

Edited by Mrnhan, 28-10-2014 - 21:56.