A. 1
B.
C.
D. 4
E.
[ 5-6224 - op net sinds 13.5.12-(E)-15.9.2025 ]
Oplossing - Solution
1
ste
manier : zonder de regel van de L'Hospital
2
de
manier : met de regel van de L'Hospital
\left(\sqrt{x+3}+2\right)}{\left(x^2-1\right)\left(\sqrt{x+3}+2\right)}\\=\lim_{x\to1}\frac{x+3-4}{\left(x^2-1\right)\left(\sqrt{x+3}+2\right)}\\=\lim_{x\to1}\frac{x\!-\!1}{\left(x\!-\!1\right)\left(x+1\right)\left(\sqrt{x+3}+2\right)}\\=\lim_{x\to1}\frac{1}{\left(x+1\right)\left(\sqrt{x+3}+2\right)}=\frac{1}{2\times4}=\frac18&space; "\\\lim_{x\to1}\frac{\sqrt{x+3}-2}{x^2-1}=\lim_{x\to1}\frac{\left(\sqrt{x+3}-2\right)\left(\sqrt{x+3}+2\right)}{\left(x^2-1\right)\left(\sqrt{x+3}+2\right)}=\lim_{x\to1}\frac{x+3-4}{\left(x^2-1\right)\left(\sqrt{x+3}+2\right)}\\=\lim_{x\to1}\frac{x\!-\!1}{\left(x\!-\!1\right)\left(x+1\right)\left(\sqrt{x+3}+2\right)}=\lim_{x\to1}\frac{1}{\left(x+1\right)\left(\sqrt{x+3}+2\right)}=\frac{1}{2\times4}=\frac18")
2de manier : met de regel van de L'Hospital
\overset{H}=\lim_{x\to1}\frac{\frac{1}{2\sqrt{x+3}}}{2x}=\frac{\frac{1}{2.2}}{2}=\frac{1}{8} "\lim_{x\to1}\frac{\sqrt{x+3}-2}{x^2-1}\left(=\frac{0}{0}\right)\overset{H}=\lim_{x\to1}\frac{\frac{1}{2\sqrt{x+3}}}{2x}=\frac{\frac{1}{2.2}}{2}=\frac{1}{8}")
