Wiskunde Multiple Choice Vraag

$\displaystyle \lim_{x \to a}\large \frac {\cos 2x - \cos 2a} {\cos x - \cos a} =$	A. 0
B. 2
C. 2. cos a
D. 4
E. 4. cos a

$\displaystyle \lim_{x \to a}\large \frac {\cos 2x - \cos 2a} {\cos x - \cos a} =$	A. 0
B. 2
C. 2(cos a)
D. 4
E. 4(cos a)

1^ste manier : met de formules van SIMPSON
$\cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$
$\inline\displaystyle\lim_{x\to a}\large\frac{\cos 2x-\cos 2a}{\cos x-\cos a}=\lim_{x\to a}\frac{-2\sin(x+a)\sin(x-a)}{-2\sin\frac{x+a}{2}.\sin\frac{x-a}{2}}\\=\frac{\sin2a}{\sin a}.\lim_{x\to a}\frac{\sin(x-a)}{\sin\frac{x-a}{2}}=\frac{2\sin a\cos a}{\sin a}.\lim_{x\to a}\frac{\sin(x-a)}{\sin\frac{x-a}{2}}\\=2\cos a.\left(\lim_{x\to a}\frac{\frac{\sin(x-a)}{x-a}}{\frac{\sin\frac{x-a}{2}}{\frac{x-a}{2}}}\right).\lim_{x\to a}\frac{x-a}{\frac{x-a}{2}}=2\cos a.\frac11.2=4\cos a$
2^de manier : met de regel van de L'Hospital
$\displaystyle\lim_{x\to a}\large\frac{\cos 2x-\cos 2a}{\cos x-\cos a}\left(=\frac00\right)\buildrel H\over=\lim_{x\to a}\frac{-2\sin 2x}{-\sin x}=\lim_{x\to a}\frac{2.2.\sin x.\cos x}{\sin x}=4.\cos x$

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