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[ 6-5537 - op net sinds .3.09-(E)-20.12.2023 ]
Translation in E N G L I S H
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Oplossing - Solution
1ste manier : door ontbinden van x² − 1 en x − 1 als een verschil van kwadraten
\left(x+1\right)}=\lim_{x\to1}\;\frac{-\left(\sqrt&space;x-1\right)}{\left(\sqrt&space;x-1\right)\left(\sqrt&space;x+1\right)\left(x+1\right)}\\=\lim_{x\to1}\;\frac{-1}{\left(\sqrt&space;x+1\right)\left(x+1\right)}=\frac{-1}{2.2}=-\;\;\frac14&space; "\\\lim_{x\to1}\;\frac{1-\sqrt x}{x^2-1}=\lim_{x\to1}\;\frac{1-\sqrt x}{\left(x-1\right)\left(x+1\right)}=\lim_{x\to1}\;\frac{-\left(\sqrt x-1\right)}{\left(\sqrt x-1\right)\left(\sqrt x+1\right)\left(x+1\right)}\\=\lim_{x\to1}\;\frac{-1}{\left(\sqrt x+1\right)\left(x+1\right)}=\frac{-1}{2.2}=-\;\;\frac14")
2de manier : met behulp van toegevoegde wortelvormen
\left(1+\sqrt&space;x\right)}{\left(x^2-1\right)\left(1+\sqrt&space;x\right)}=\lim_{x\to1}\;\frac{1-x}{\left(x-1\right)\left(x+1\right)\left(1+\sqrt&space;x\right)}\\=\lim_{x\to1}\;\frac{-1}{\left(x+1\right)\left(1+\sqrt&space;x\right)}=\frac{-1}{2.2}=-\;\;\frac14 "\\\lim_{x\to1}\;\frac{1-\sqrt x}{x^2-1}=\lim_{x\to1}\;\frac{\left(1-\sqrt x\right)\left(1+\sqrt x\right)}{\left(x^2-1\right)\left(1+\sqrt x\right)}=\lim_{x\to1}\;\frac{1-x}{\left(x-1\right)\left(x+1\right)\left(1+\sqrt x\right)}\\=\lim_{x\to1}\;\frac{-1}{\left(x+1\right)\left(1+\sqrt x\right)}=\frac{-1}{2.2}=-\;\;\frac14")
3de manier : met behulp van de regel van de l'Hospital
\overset{H}{=}\;\lim_{x\to1}\;\frac{-\frac{1}{2\sqrt&space;x}}{2x}=-\;\;\lim_{x\to1}\;\frac{1}{4\sqrt&space;x}=-\;\;\frac14 "\lim_{x\to1}\;\frac{1-\sqrt x}{x^2-1}\left(=\frac{0}{0}\right)\overset{H}{=}\;\lim_{x\to1}\;\frac{-\frac{1}{2\sqrt x}}{2x}=-\;\;\lim_{x\to1}\;\frac{1}{4\sqrt x}=-\;\;\frac14")
