[ 5-6223 - op net sinds 23.3.12-(E)-3.2.2024 ]
Translation in E N G L I S H
Oplossing - Solution
1ste manier : zonder de regel van de l'Hospital
=1.1.1.1.4=4&space; "\\\displaystyle\lim_{x\to 0}\;\frac{\large 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 x.\sin x}{\sin\frac x2.\sin\frac x2}\\=\lim_{x\to 0}\;\left(\frac{\sin x}{x}.\frac{\sin x}{x}.\frac{\frac x2}{\sin\frac x2}.\frac{\frac x2}{\sin\frac x2}.4\right)=1.1.1.1.4=4")
2de manier : zonder de regel van de l'Hospital
}{1-\cos&space;x}\\\\=2\lim_{x\to&space;0}\;(1+\cos&space;x)=2.(1+1)=4&space; "\\\displaystyle\lim_{x\to 0}\;\frac{\large 1\,-\,\cos 2x}{1\,-\,\cos x}=\lim_{x\to 0}\;\frac{2\sin^2x}{1-\cos x}=\lim_{x\to 0}\;\frac{2(1-\cos^2x)}{1-\cos x}\\\\=2\lim_{x\to 0}\;(1+\cos x)=2.(1+1)=4")
3de manier : met de regel van de L'Hospital
\buildrel&space;H\over=\lim_{x\to&space;0}\;\frac{2\sin2x}{\sin&space;x}\\=\lim_{x\to&space;0}\;\frac{4\sin&space;x\cos&space;x}{\sin&space;x}\lim_{x\to&space;0}\;(4.\cos&space;x)=4.1=4&space; "\\\displaystyle\lim_{x\to 0}\;\frac{\large 1\,-\,\cos 2x}{1\,-\,\cos x}\left(=\frac00\right)\buildrel H\over=\lim_{x\to 0}\;\frac{2\sin2x}{\sin x}\\=\lim_{x\to 0}\;\frac{4\sin x\cos x}{\sin x}\lim_{x\to 0}\;(4.\cos x)=4.1=4")
4de manier : met de regel van de L'Hospital
\buildrel&space;H\over=\lim_{x\to&space;0}\;\frac{2\sin2x}{\sin&space;x}\\=2.\lim_{x\to&space;0}\;\left(\frac{\sin&space;2x}{2x}.\frac{x}{\sin&space;x}\right).2=2.1.1.2=4&space; "\\\displaystyle\lim_{x\to 0}\;\frac{\large 1\,-\,\cos 2x}{1\,-\,\cos x}\left(=\frac00\right)\buildrel H\over=\lim_{x\to 0}\;\frac{2\sin2x}{\sin x}\\=2.\lim_{x\to 0}\;\left(\frac{\sin 2x}{2x}.\frac{x}{\sin x}\right).2=2.1.1.2=4")
