Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\lim_{x \to \, -2}\, \frac{x^3+8}{x^2-4} =$	A. − 1
B. − 2
C. − 3
D. − 4
E.

$\\\lim_{x \to \, 2} \frac{x^3\!+8}{x^2\!-4} \!=\!\lim_{x \to \, 2} \frac {(x+2)(x^2\!-2x+\!4)} {(x-\!2)(x+2)} \!=\!\lim_{x \to \, 2} \frac {x^2\!-\!2x+\!4} {x-2}\! =\!\frac{4\!+\!4\!+\!4}{-2-2} =\!\frac{12}{-4}\!=\!-3$