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[ 5-7064 - op net sinds 21.8.2025-(E)-16.4.2026 ]
Translation in E N G L I S H
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Oplossing - Solution
1ste manier :met de regel van de l'Hospital'
\overset{H}{=}\lim_{x\to\frac{\pi}{2}}\frac{-\sin&space;x}{-5\sin5x}\\&=\frac15\lim_{x\to\frac{\pi}{2}}\frac{\sin&space;x}{\sin5x}=\frac15.\frac11=\frac15&space;\end{aligned} "\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{\cos x}{\cos 5x}(=\frac00)\overset{H}{=}\lim_{x\to\frac{\pi}{2}}\frac{-\sin x}{-5\sin5x}=\frac15\lim_{x\to\frac{\pi}{2}}\frac{\sin x}{\sin5x}=\frac15.\frac11=\frac15")
2de manier : zonder de regel van de l'Hospital'
}{\sin\left(\frac{\pi}{2}-5x\right)}\\&=\lim_{x\to\frac{\pi}2}\frac{\sin\left(\frac{\pi}{2}-x\right)}{\sin\left(\frac{5\pi}{2}-5x\right)}\\&=\lim_{x\to\frac{\pi}2}\frac{\left(\frac{5\pi}2-5x\right).\sin\left(\frac{\pi}2-x\right).\left(\frac{\pi}{2}-x\right)}{\sin\left(\frac{5\pi}2-5x\right).\left(\frac{\pi}{2}-x\right).\left(\frac{5\pi}{2}-5x\right)}\\&=\lim_{x\to\frac{\pi}2}\frac{\frac{\pi}{2}-x}{\frac{5\pi}{2}-5x}=\lim_{x\to\frac{\pi}2}\frac{\frac{\pi}{2}-x}{5.\left(\frac{\pi}{2}-x\right)}=\frac15\end{aligned} "\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{\cos x}{\cos 5x}=\lim_{x\to\frac{\pi}2}\frac{\sin\left(\frac{\pi}2-x\right)}{\sin\left(\frac{\pi}{2}-5x\right)}=\lim_{x\to\frac{\pi}2}\frac{\sin\left(\frac{\pi}{2}-x\right)}{\sin\left(\frac{5\pi}{2}-5x\right)}=\lim_{x\to\frac{\pi}2}\frac{\left(\frac{5\pi}2-5x\right).\sin\left(\frac{\pi}2-x\right).\left(\frac{\pi}{2}-x\right)}{\sin\left(\frac{5\pi}2-5x\right).\left(\frac{\pi}{2}-x\right).\left(\frac{5\pi}{2}-5x\right)}=\lim_{x\to\frac{\pi}2}\frac{\frac{\pi}{2}-x}{\frac{5\pi}{2}-5x}\\=\lim_{x\to\frac{\pi}2}\frac{\frac{\pi}{2}-x}{5.\left(\frac{\pi}{2}-x\right)}=\frac15")
