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[ 5-8656 - op net sinds 3.5.2021-(E)-31.1.2024 ]
Translation in E N G L I S H
see above
Oplossing - Solution
1
ste
manier : zonder de regel van de l' Hopital
2
de
manier : met de regel van de l' Hospital
}{1-2.\cos&space;x}\qquad\qquad\text{(subst.}\;\;x-\frac{\pi}{3}=t)\\=\lim_{{x\to\frac{\pi}{3}}}\;\frac{\sin&space;t}{1-2.\cos(t+\frac{\pi}{3})}\qquad\qquad\overset{t\,\rightarrow\;0\;\;als\;\;x\rightarrow\frac{\pi}{3}}{}\\=\lim_{{x\to\frac{\pi}{3}}}\;\frac{\sin&space;t}{1-2.(\cos&space;t.\cos\frac{\pi}{3}-\sin&space;t.\sin\frac{\pi}{3})} "\\\displaystyle\lim_{{x\to\frac{\pi}{3}}}\;\frac{\sin(x-\frac{\pi}{3})}{1-2.\cos x}\qquad\qquad\text{(subst.}\;\;x-\frac{\pi}{3}=t)\\=\lim_{{x\to\frac{\pi}{3}}}\;\frac{\sin t}{1-2.\cos(t+\frac{\pi}{3})}\qquad\qquad\overset{t\,\rightarrow\;0\;\;als\;\;x\rightarrow\frac{\pi}{3}}{}\\=\lim_{{x\to\frac{\pi}{3}}}\;\frac{\sin t}{1-2.(\cos t.\cos\frac{\pi}{3}-\sin t.\sin\frac{\pi}{3})}")
2de manier : met de regel van de l' Hospital
}{1-2.\cos&space;x}=\left(=\frac00\right)\\\buildrel&space;H\over=\lim_{{x\to\frac{\pi}{3}}}\;\frac{\cos(x-\frac{\pi}{3})}{2.\sin&space;x}=\frac{\cos0}{2.sin\frac{\pi}{3}}\\=\frac{1}{2.\frac{\sqrt3}{2}}=\frac{1}{\sqrt3}=\frac{\sqrt3}{3} "\\\displaystyle\lim_{{x\to\frac{\pi}{3}}}\;\frac{\sin(x-\frac{\pi}{3})}{1-2.\cos x}=\left(=\frac00\right)\\\buildrel H\over=\lim_{{x\to\frac{\pi}{3}}}\;\frac{\cos(x-\frac{\pi}{3})}{2.\sin x}=\frac{\cos0}{2.sin\frac{\pi}{3}}\\=\frac{1}{2.\frac{\sqrt3}{2}}=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}")
