Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\displaystyle \lim_{x \to \frac{\pi}{3}}\; \frac {\sin \frac{\pi}{3}\,-\,\sin x} {\cos\frac{\pi}{3}\,-\,\cos x}$ is gelijk aan	A. −
B.
C.
D. −
E.

$\displaystyle \lim_{x \to \frac{\pi}{3}}\; \frac {\sin \frac{\pi}{3}\,-\,\sin x} {\cos\frac{\pi}{3}\,-\,\cos x}$ is equal to	A. −
B.
C.
D. −
E.

1^ste manier :
$\\\displaystyle\lim_{x\to\frac{\pi}{3}}\;\frac{\sin\frac{\pi}{3}\,-\,\sin x}{\cos\frac{\pi}{3}\,-\,\cos x}\overset{SIMPSON}{=}\displaystyle\lim_{x\to\frac{\pi}{3}}\;\frac{2\cos\left(\frac{\pi}{6}+\frac x2\right).\sin\left(\frac{\pi}{6}-\frac x2\right)}{-2\sin\left(\frac{\pi}{6}+\frac x2\right).\sin\left(\frac{\pi}{6}-\frac x2\right)}=-\displaystyle\lim_{x\to\frac{\pi}{3}}\;\frac{\cos\left(\frac{\pi}{6}+\frac x2\right)}{\sin\left(\frac{\pi}{6}+\frac x2\right)}\\=-\displaystyle\lim_{x\to\frac{\pi}{3}}\;\cot\left(\frac{\pi}{6}+\frac{x}{2}\right)=\cot\left(\frac{\pi}{6}+\frac{\pi}{6}\right)=-\cot\frac{\pi}{3}=-\frac{\sqrt3}{3}$
2^de manier :
$\\\displaystyle\lim_{x\to\frac{\pi}{3}}\frac{\sin\frac{\pi}{3}-\sin x}{\cos\frac{\pi}{3}\!-\cos x}\left(\!=\frac00\right)\buildrel H\over=\displaystyle\lim_{x\to\frac{\pi}{3}}\frac{-\cos x}{\sin x}\!=\!-\!\displaystyle\lim_{x\to\frac{\pi}{3}}\cot x\!=\!-\!\cot\frac{\pi}{3}\!=\!-\!\cot 60^\circ\!=\!-\frac{\sqrt3}{3}$