A. 1
B. 2
C.
D.
E. 0
[ 5-A164 - op net sinds 31.3.2026-(E)-12.4.2026 ]
Translation in E N G L I S H
See above
Oplossing - Solution
1
ste
manier : zonder de regel van de L'Hospital
2
de
manier : met de regel van de L'Hospital
}{(1-\cos&space;x).(1+\cos&space;x)}\\=\lim_{\underset{<}{x\to\,0}}\;\frac{\sin&space;x\cdot(1+\cos&space;x)}{1-\cos^2x}=\lim_{\underset{<}{x\to\,0}}\;\frac{\sin&space;x\cdot(1+\cos&space;x)}{\sin^2x}\\=\lim_{\underset{<}{x\to\,0}}\;\frac{1+\cos&space;x}{\sin&space;x}=\frac{2}{'-0'}=-\infty&space; "\inline\displaystyle\lim_{\underset{<}{x\to\,0}}\;\frac{\sin x}{1-\cos x}=\lim_{\underset{<}{x\to\,0}}\;\frac{\sin x\cdot(1+\cos x)}{(1-\cos x).(1+\cos x)}\\=\lim_{\underset{<}{x\to\,0}}\;\frac{\sin x\cdot(1+\cos x)}{1-\cos^2x}=\lim_{\underset{<}{x\to\,0}}\;\frac{\sin x\cdot(1+\cos x)}{\sin^2x}\\=\lim_{\underset{<}{x\to\,0}}\;\frac{1+\cos x}{\sin x}=\frac{2}{'-0'}=-\infty")
2de manier : met de regel van de L'Hospital
\overset{H}{=}\lim_{\underset{<}{x\to\,0}}\;\frac{\cos&space;x}{\sin&space;x}\\=\lim_{\underset{<}{x\to\,0}}\;\cot&space;x=-\infty&space; "\inline\displaystyle\\\lim_{\underset{<}{x\to\,0}}\;\frac{\sin x}{1-\cos x}\left(=\frac00\right)\overset{H}{=}\lim_{\underset{<}{x\to\,0}}\;\frac{\cos x}{\sin x}\\=\lim_{\underset{<}{x\to\,0}}\;\cot x=-\infty")
