Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

Bereken de volgende som : $https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \frac {\cos 3x} {\cos x} +\displaystyle \lim_{x \to \frac{\pi }{2}} \frac {\cos 3x} {\cos x}$	A. − 8
B. − 2
C. 0
D. 4
E. een niet geheel getal

Evaluate : $https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \frac {\cos 3x} {\cos x} +\displaystyle \lim_{x \to \frac{\pi }{2}} \frac {\cos 3x} {\cos x}$	A. − 8
B. − 2
C. 0
D. 4
E. none of these

1^ste manier : met de regel van de L'Hospital (applying L'Hospital rule)
$\\\displaystyle\lim_{x\to 0}\frac{\cos 3x}{\cos x}+\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{\cos 3x}{\cos x}=\frac11.\left(=\frac00\right)\buildrel H\over=1+\displaystyle\lim_{x\to 0}\frac{-3\sin3x}{-\sin x}\\=1+3.\frac{\sin\frac{3\pi}{2}}{\sin\frac{\pi}{2}}=1+3.\frac{-1}{1}=1-3=-2$
2^de manier : zonder de regel van de L'Hospital (without L'Hospital rule)
$\\\displaystyle\lim_{x\to 0}\frac{\cos 3x}{\cos x}+\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{\cos 3x}{\cos x}=\frac11+\displaystyle\lim_{x\to\frac{\pi}2}\frac{\sin\left(\frac{\pi}2-3x\right)}{\sin\left(\frac{\pi}{2}-x\right)}=1+\displaystyle\lim_{x\to\frac{\pi}2}\frac{-\sin\left(\frac{3\pi}{2}-3x\right)}{\sin\left(\frac{\pi}{2}-x\right)}\\=1-\displaystyle\lim_{x\to\frac{\pi}2}\frac{\left(\frac{\pi}2-x\right).\sin\left(\frac{3\pi}2-3x\right).\left(\frac{3\pi}{2}-3x\right)}{\sin\left(\frac{\pi}2-x\right).\left(\frac{3\pi}{2}-3x\right).\left(\frac{\pi}{2}-x\right)}=1-\displaystyle\lim_{x\to\frac{\pi}2}\frac{\frac{3\pi}{2}-3x}{\frac{\pi}{2}-x}\\=1-\displaystyle\lim_{x\to\frac{\pi}2}\frac{3.\left(\frac{\pi}{2}-x\right)}{\frac{\pi}{2}-x}=1-3=-2$