Wiskunde Multiple Choice Vraag

$\inline\\\displaystyle\lim_{x\to\,0}\;\frac{\cos 2x-1}{\cos x-1}=$	A. 1
B. 2
C. 4
D.
E.

$\inline\\\displaystyle\lim_{x\to\,0}\;\frac{\cos 2x-1}{\cos x-1}=\lim_{x\to\,0}\;\frac{1-\cos 2x}{1-\cos x}=\lim_{x\to\,0}\;\frac{2\sin^2x}{2\sin^2\frac x2}=\lim_{x\to\,0}\;\frac{\sin^2x}{\sin^2\frac x2}\\\\=\lim_{x\to\,0}\;\left(\frac{\sin x}{\sin\frac x2}\right)^2=\lim_{x\to\,0}\;\left(\frac{\sin x}{x}\cdot\frac{2.\frac x2}{\sin\frac x2}\right)^2=(1.2.1)^2=4$

^{gricha - vA160 - 30.3.2026}