Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\displaystyle\lim_{x\to\,0}\frac{x^6-\sin^6x}{x^8}$	A. 1
B. 2
C. \(\frac34\)
D.
E.

$\inline\\\displaystyle\lim_{x\to\,0}\frac{x^6-\sin^6x}{x^8}=\lim_{x\to\,0}\left[\frac{x^3-\sin^3x}{x^5}\cdot\frac{x^3+\sin^3x}{x^3}\right]\\=\lim_{x\to\,0}\left[\frac{x-\sin x}{x^3}\cdot\frac{x^2+x.\sin x+\sin^2x}{x^2}\cdot\frac{x^3+\sin^3x}{x^3}\right]\\=\lim_{x\to\,0}\left[\frac{x-\sin x}{x^3}\cdot\left(1+\frac{\sin x}{x}+\frac{\sin^2x}{x^2}\right)\cdot\left(1+\frac{\sin^3x}{x^3}\right)\right]\\\left[\lim_{x\to\,0}\frac{x-\sin x}{x^3}\right].(1+1+1)(1+1)\\=6\cdot\lim_{x\to\,0}\frac{x-\sin x}{x^3}\overset{H}{=}6.\lim_{x\to\,0}\frac{1-\cos x}{3x^2}\overset{H}{=}6\cdot\frac{\sin x}{6x}=6\cdot\frac16=1$