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[ 5-6624 - op net sinds 7.4.12-(E)-26.9.2024 ]
Translation in E N G L I S H
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Oplossing - Solution
1ste manier :
+x+1}{x^2-1} "\\\lim_{x\to-1}\;\frac{x^3+1}{x^2-1}=\lim_{x\to-1}\;\frac{x^3-x+x+1}{x^2-1}=\lim_{x\to-1}\;\frac{x(x^2-1)+x+1}{x^2-1}\\=\lim_{x\to-1}\;x+\lim_{x\to-1}\;\frac{x+1}{(x+1)(x-1)}=\lim_{x\to-1}\;x+\lim_{x\to-1}\;\frac{1}{x-1}=-1+\frac{1}{-2}\\=-1-\frac12=-\frac32")
(x-1)}=\lim_{x\to-1}\;x+\lim_{x\to-1}\;\frac{1}{x-1}=-1+\frac{1}{-2}\\=-1-\frac12=-\frac32 "\\\lim_{x\to-1}\;\frac{x^3+1}{x^2-1}=\lim_{x\to-1}\;\frac{x^3-x+x+1}{x^2-1}=\lim_{x\to-1}\;\frac{x(x^2-1)+x+1}{x^2-1}\\=\lim_{x\to-1}\;x+\lim_{x\to-1}\;\frac{x+1}{(x+1)(x-1)}=\lim_{x\to-1}\;x+\lim_{x\to-1}\;\frac{1}{x-1}=-1+\frac{1}{-2}\\=-1-\frac12=-\frac32")
2de manier : teller en noemer ontbinden in factoren
(x^2\!-\!x\!+\!1)}{(x+1)(x-1)}=\!\lim_{x\to-1}\,\frac{x^2\!-\!x\!+\!1}{x-1}=\frac{1\!+1\!+\!1}{-1-1}=-\frac32 "\\\lim_{x\to-1}\,\frac{x^3+1}{x^2-1}=\!\lim_{x\to-1}\,\frac{(x\!+\!1)(x^2\!-\!x\!+\!1)}{(x+1)(x-1)}=\!\lim_{x\to-1}\,\frac{x^2\!-\!x\!+\!1}{x-1}=\frac{1\!+1\!+\!1}{-1-1}=-\frac32")
3de manier : me de regel van de l'Hospital
\overset{H}{=}\lim_{x\to-1}\;\frac{3x^2}{2x}=\lim_{x\to-1}\,\frac{3x}{2}=-\frac32 "\\\lim_{x\to-1}\;\frac{x^3+1}{x^2-1}\left(=\frac00\right)\overset{H}{=}\lim_{x\to-1}\;\frac{3x^2}{2x}=\lim_{x\to-1}\,\frac{3x}{2}=-\frac32")
