Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\displaystyle \lim_{x \to 0}\; \frac {1-\,\cos 2x.\cos x} {1\:-\:\cos x}$ is gelijk aan	A. 1
B. 2
C. 3
D. 4
E. 5

$\displaystyle \lim_{x \to 0}\; \frac {1-\,\cos 2x.\cos x} {1\:-\:\cos x}$ is equal to	A. 1
B. 2
C. 3
D. 4
E. 5

1^ste manier : zonder de regel van de l’ Hospital
gebruik makend van cos 2x = 1 − 2sin²x en ontbinden in factoren
$\inline\\\displaystyle\lim_{x\to\,0}\;\frac{1-\,\cos 2x.\cos x}{1\:-\:\cos x}=\lim_{x\to\,0}\;\frac{1-(1-2\sin^2x).\cos x}{1-\cos x}\\=\lim_{x\to\,0}\;\frac{1-\cos x+2\sin^2x.\cos x}{1-\cos x}\\=\lim_{x\to\,0}\;\frac{1-\cos x+2(1-\cos^2x).\cos x}{1-\cos x}\\=\lim_{x\to\,0}\;\frac{(1-\cos x)\left(1+2(1+\cos x).\cos x\right)}{1-\cos x}\\=\lim_{x\to\,0}\;\left(1+2(1+\cos x).\cos x\right)=1+2(1+1).1=1+4=5$
2^de manier : zonder de regel van de l’ Hospital
gebruik makend van cos 2x = 2cos² x − 1 en met susbtitutie y = cos x
$\inline\\\displaystyle\lim_{x\to\,0}\;\frac{1-\,\cos 2x.\cos x}{1\:-\:\cos x}=\lim_{x\to\,0}\;\frac{1-(2\cos^2x-1).\cos x}{1-\cos x}\\=\lim_{x\to\,0}\;\frac{-2\cos^2x+\cos x+1}{1-\cos x}=(*)\\Stel\;\cos x=y\;\Rightarrow\;als\;x\to 0\;dan\;y\to 1\\(*)=\lim_{y\to\,1}\;\frac{-2y^2+y+1}{1-y}=\lim_{y\to\,1}\;\frac{(y-1)(-2y^2-2y-1)}{1-y}\\=\lim_{y\to\,1}\;(2y^2+2y+1)=2.1^2+2.1+1=2+2+1=5$
3^de manier : zonder de regel van de l’ Hospital
teller en noemer vermenigvuldigen met 1 + cos x
$\inline\\\displaystyle\lim_{x\to\,0}\;\frac{1-\,\cos 2x.\cos x}{1\:-\:\cos x}\lim_{x\to\,0}\;\frac{(1-\,\cos 2x.\cos x)(1+\cos x)}{(1\:-\:\cos x)(1+\cos x)}\\=\lim_{x\to\,0}\;\frac{1+\cos x-\cos 2x.\cos x-\cos 2x.\cos^2x}{1-\cos^2x}\\=\lim_{x\to\,0}\;\frac{1+\cos x-(1-2\sin^2x)\cos x-(1-2\sin^2x)(1-\sin^2x)}{\sin^2x}\\=\lim_{x\to\,0}\;\frac{1+\cos x-\cos x+2\sin^2x.\cos x-1+\sin^2x+2\sin^2x-2\sin^3x}{\sin^2x}\\\lim_{x\to\,0}\;\frac{2\sin^2x.\cos x+\sin^2x+2\sin^2x-2\sin^3x}{\sin^2x}\\\lim_{x\to\,0}\;(2\cos x+1+2-\sin x)=2.1+1+2-0=5$ 4^de manier : met de regel van de l’ Hospital
$\inline\\\displaystyle\lim_{x\to\,0}\;\frac{1-\,\cos 2x.\cos x}{1\:-\:\cos x}\left(=\frac00\right)\overset{H}{=}\lim_{x\to\,0}\;\frac{0+2\sin 2x.\cos x+\cos 2x.\sin x}{\sin x}\\\lim_{x\to\,0}\;(4\cos^2x+\cos 2x)=4.1+1=5$

^{gricha - v7100 - 25.4.2026}