Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

Bereken de volgende limiet (van een product) $\displaystyle \lim_{x \to \frac{\pi }{3}}\; \left ( \sin \frac{\pi}{3} - \sin x \right).\tan \frac{3x}{2}$	A.
B.
C.
D.
E.

Compute the following limit ( of a product) $\displaystyle \lim_{x \to \frac{\pi }{3}}\; \left ( \sin \frac{\pi}{3} - \sin x \right).\tan \frac{3x}{2}$	A.
B.
C.
D.
E.

1^ste manier : zonder de regel van de l'Hospital
$\\\lim_{x\to\frac{\pi}{3}}\left(\sin\frac{\pi}{3}-\sin x\right).\tan\frac{3x}{2}\overset{SIMPSON}{=}\lim_{x\to\frac{\pi}{3}}\frac{\sin\frac{3x}{2}.2\cos\left(\frac{\pi}{6}+\frac x2\right)\sin\left(\frac{\pi}{6}-\frac x2\right)}{\cos\frac{3x}{2}}\\=\lim_{x\to\frac{\pi}{3}}\frac{2\sin\frac{3x}{2}\cos\left(\frac{\pi+3x}{6}\right)\sin\left(\frac{\pi-3x}{6}\right)}{\cos\frac{3x}{2}}=2.1.\frac12.\lim_{x\to\frac{\pi}{2}}\frac{\sin\left(\frac{\pi-3x}{6}\right)}{\cos\frac{3x}{2}}\\=\lim_{x\to\frac{\pi}{3}}\frac{\sin\left(\frac{\pi-3x}{6}\right)}{\sin\left(\frac{\pi-3x}{2}\right)}=\lim_{x\to\frac{\pi}{3}}\frac{\left(\frac{\pi-3x}{6}\right).\frac{\sin\left(\frac{\pi-3x}{6}\right)}{\frac{\pi-3x}{6}}}{\left(\frac{\pi-3x}{2}\right).\frac{\sin\left(\frac{\pi-3x}{2}\right)}{\frac{\pi-3x}{2}}}=\lim_{x\to\frac{\pi}{3}}\frac{\left(\frac{\pi-3x}{6}\right).1}{\left(\frac{\pi-3x}{2}\right).1}=\frac26=\frac13$
2^de manier : met zonder de regel van de l'Hospital
$\\\lim_{x\to\frac{\pi}{3}}\left(\sin\frac{\pi}{3}-\sin x\right).\tan\frac{3x}{2}=\lim_{x\to\frac{\pi}{3}}\frac{\sin\frac{\pi}{3}-\sin x}{\cot\frac{3x}{2}}\left(=\frac00\right)\\\overset{H}{=}\lim_{x\to\frac{\pi}{3}}\frac{0-cos x}{-\frac{1}{\sin^2\frac{3x}{2}}.\frac32}=\frac{\frac12}{\frac{1}{1^2}\frac32}=\frac12.\frac23=\frac13$