A. 1
B. π
C. −π
D. 2π
E. 4π
[ 5-139 - op net sinds 2.3.2026-(E)- ]
Translation in E N G L I S H
A. 1
B. π
C. −π
D. 2π
E. 4π
Oplossing - Solution
1
ste
manier : met de regel van de l'Hospital'
2
de
manier : zonder de regel van de l'Hospital
}{x^2}= "\displaystyle\lim_{x\to\,0}\frac{\sin(\pi\cdot\cos^2x)}{x^2}=")
}{x^2}= "\displaystyle\lim_{x\to\,0}\frac{\sin(\pi\cdot\cos^2x)}{x^2}=")
}{x^2}\left(=\frac00\right)\\\overset{H}{=}\lim_{x\to\,0}\frac{\cos(\pi\cdot\cos^2x)\cdot\pi\cdot2\cdot\cos&space;x\cdot(-\sin&space;x)}{2x}\\=\lim_{x\to\,0}\left(\cos(\pi\cdot\cos^2x)\cdot\pi\cdot\cos&space;x\;\right)\cdot\lim_{x\to\,0}\left(-\,\frac{\sin&space;x}x\right)\\=(\cos\pi)\cdot\pi\cdot\cos&space;0.(-1)=\pi "\inline\\\displaystyle\lim_{x\to\,0}\frac{\sin(\pi\cdot\cos^2x)}{x^2}\left(=\frac00\right)\\\overset{H}{=}\lim_{x\to\,0}\frac{\cos(\pi\cdot\cos^2x)\cdot\pi\cdot2\cdot\cos x\cdot(-\sin x)}{2x}\\=\lim_{x\to\,0}\left(\cos(\pi\cdot\cos^2x)\cdot\pi\cdot\cos x\;\right)\cdot\lim_{x\to\,0}\left(-\,\frac{\sin x}x\right)\\=(\cos\pi)\cdot\pi\cdot\cos 0.(-1)=\pi")
2de manier : zonder de regel van de l'Hospital
}{x^2}=\lim_{x\to\,0}\left[\frac{\sin(\pi-\pi\cos^2x)}{\pi-\pi\cdot\cos^2x}\cdot\frac{\pi-\pi\cdot\cos^2x}{x^2}\right]\\=1.\lim_{x\to\,0}\frac{\pi-\pi\cdot\cos^2x}{x^2}=\pi\lim_{x\to\,0}\frac{1-\cos^2x}{x^2}=\pi\lim_{x\to\,0}\frac{\sin^2x}{x^2}=\pi.1^2=\pi "\inline\\\displaystyle\lim_{x\to\,0}\frac{\sin(\pi\cdot\cos^2x)}{x^2}=\lim_{x\to\,0}\left[\frac{\sin(\pi-\pi\cos^2x)}{\pi-\pi\cdot\cos^2x}\cdot\frac{\pi-\pi\cdot\cos^2x}{x^2}\right]\\=1.\lim_{x\to\,0}\frac{\pi-\pi\cdot\cos^2x}{x^2}=\pi\lim_{x\to\,0}\frac{1-\cos^2x}{x^2}=\pi\lim_{x\to\,0}\frac{\sin^2x}{x^2}=\pi.1^2=\pi")
