Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\large \dpi{110}\displaystyle \lim_{x \to \frac{\pi }{4}}\; (\tan 2x.\ln \tan x)$ is gelijk aan	A. − 4
B. − 1
C.
D. 0
E.

$https://latex.codecogs.com/png.image?\large \dpi{110}\displaystyle \lim_{x \to \frac{\pi }{4}}\; (\tan 2x.\ln \tan x)$ is equal to	A. − 4
	B. − 1
	C.
	D. 0
	E. ∞

$\ \\displaystyle\lim_{x\to\frac{\pi}{4}}\;(\tan 2x.\ln\tan x)=\displaystyle\lim_{x\to\frac{\pi}{4}}\;\frac{\ln\tan x}{\cot 2x}\left(=\frac00\right)\overset{H}{=} \displaystyle\lim_{x\to\frac{\pi}{4}}\;\frac{\frac{1}{\tan x}.\frac{1}{\cos^2x}}{-\frac{1}{\sin^2 2x}.2}= -\displaystyle\lim_{x\to\frac{\pi}{4}}\;\frac{\frac{1}{\sin x}.\frac{1}{\cos x}}{\frac{2}{\sin^2 2x}}\\= -\displaystyle\lim_{x\to\frac{\pi}{4}}\;\frac{\sin^22x}{2.\sin x.\cos x}= -\displaystyle\lim_{x\to\frac{\pi}{4}}\;\frac{\sin^22x}{\sin 2x}= -\displaystyle\lim_{x\to\frac{\pi}{4}}\;\sin 2x=-\sin\frac{\pi}{2}=-1$