[ 5-8732 - op net sinds 7.4.2021-(E)-16.1.2024 ]
Translation in E N G L I S H
Oplossing - Solution
1ste manier : d.m.v. de substitutie \(1-x=\frac{2t}{\pi}\)
⇔ \(x=1-\frac{2t}{\pi}\) ⇔ \(\frac{\pi.x}{2}=\frac{\pi}2-t\)
Als x → 1 dan t → 0
.\tan\frac{\pi&space;x}{2}\right]=\displaystyle\lim_{t\to&space;0}\;\left[\frac{2t}{\pi}.\tan\left(\frac{\pi}{2}-t\right)\right]\\=\frac{2}{\pi}.\displaystyle\lim_{t\to&space;0}\;(t.\cos&space;t)=\frac{2}{\pi}\displaystyle\lim_{t\to&space;0}\;\left(\frac{t}{\sin&space;t}.\cos&space;t\right)=\frac{2}{\pi}.1.1=\frac2{\pi} "\displaystyle\lim_{x\to 1}\;\left[(1-x).\tan\frac{\pi x}{2}\right]=\displaystyle\lim_{t\to 0}\;\left[\frac{2t}{\pi}.\tan\left(\frac{\pi}{2}-t\right)\right]\\=\frac{2}{\pi}.\displaystyle\lim_{t\to 0}\;(t.\cos t)=\frac{2}{\pi}\displaystyle\lim_{t\to 0}\;\left(\frac{t}{\sin t}.\cos t\right)=\frac{2}{\pi}.1.1=\frac2{\pi}")
2de manier : d.m.v. de regel van de l'Hospital
.\tan\frac{\pi&space;x}{2}\right]=\displaystyle\lim_{x\to&space;1}\;\frac{1-x}{\cot\frac{\pi&space;x}{2}}\left(=\frac00\right)\buildrel&space;H\over=\displaystyle\lim_{x\to&space;1}\;\frac{-1}{-\frac{1}{\sin^2\frac{\pi&space;x}{2}}.\frac{\pi}{2}}\\=\frac{2}{\pi}\displaystyle\lim_{x\to&space;1}\;\sin^2\frac{\pi&space;x}{2}=\frac{2}{\pi}.1^2=\frac{2}{\pi} "\\\displaystyle\lim_{x\to 1}\;\left[(1-x).\tan\frac{\pi x}{2}\right]=\displaystyle\lim_{x\to 1}\;\frac{1-x}{\cot\frac{\pi x}{2}}\left(=\frac00\right)\buildrel H\over=\displaystyle\lim_{x\to 1}\;\frac{-1}{-\frac{1}{\sin^2\frac{\pi x}{2}}.\frac{\pi}{2}}\\=\frac{2}{\pi}\displaystyle\lim_{x\to 1}\;\sin^2\frac{\pi x}{2}=\frac{2}{\pi}.1^2=\frac{2}{\pi}")
