1^ste manier : door ontbinden van x² − 1 en x − 1 als een verschil van kwadraten
\(\displaystyle\lim_{x\to1}\;\frac{1-\sqrt x}{x^2-1}=\lim_{x\to1}\;\frac{1-\sqrt x}{\left(x-1\right)\left(x+1\right)}=\lim_{x\to1}\;\frac{-\left(\sqrt x-1\right)}{\left(\sqrt x-1\right)\left(\sqrt x+1\right)\left(x+1\right)}\\=\displaystyle\lim_{x\to1}\;\frac{-1}{\left(\sqrt x+1\right)\left(x+1\right)}=\frac{-1}{2.2}=-\;\;\frac14\)
2^de manier : met behulp van toegevoegde wortelvormen
\(\displaystyle\lim_{x\to1}\;\frac{1-\sqrt x}{x^2-1}=\lim_{x\to1}\;\frac{\left(1-\sqrt x\right)\left(1+\sqrt x\right)}{\left(x^2-1\right)\left(1+\sqrt x\right)}=\lim_{x\to1}\;\frac{1-x}{\left(x-1\right)\left(x+1\right)\left(1+\sqrt x\right)}\\=\displaystyle\lim_{x\to1}\;\frac{-1}{\left(x+1\right)\left(1+\sqrt x\right)}=\frac{-1}{2.2}=-\;\;\frac14\)
3^de manier : met behulp van de regel van de l'Hospital
\(\displaystyle\lim_{x\to1}\;\frac{1-\sqrt x}{x^2-1}\left(=\frac{0}{0}\right)\overset{H}{=}\;\lim_{x\to1}\;\frac{-\frac{1}{2\sqrt x}}{2x}=-\;\;\lim_{x\to1}\;\frac{1}{4\sqrt x}=-\;\;\frac14\)