Wiskunde Multiple Choice Vraag

$\displaystyle\lim_{x\to\,0}\;\frac{x^6-\tan^6x}{x^8}=$	A. 1
B. − 2
C. \(\boldsymbol{\frac34}\)
D. \(\boldsymbol{\frac18}\)
E. 6

$\inline\displaystyle\lim_{x\to\,0}\;\frac{x^6-\tan^6x}{x^8}=\lim_{x\to\,0}\;\frac{x^3-\tan^3x}{x^5}\cdot\lim_{x\to\,0}\;\frac{x^3+\tan^3x}{x^3}\\=\lim_{x\to\,0}\,\frac{x^3-\tan^3x}{x^5}\cdot\left(\lim_{x\to\,0}\:\frac{x^3}{x^3}+\lim_{x\to\,0}\:\frac{\tan^3x}{x^3}\right)=2.\lim_{x\to\,0}\,\frac{x^3-\tan^3x}{x^5}\\=2\cdot\lim_{x\to\,0}\;\frac{x-\tan x}{x^3}\cdot\lim_{x\to\,0}\;\frac{x^2+x\tan x+\tan^2x}{x^2}=6\cdot\lim_{x\to\,0}\;\frac{x-\tan x}{x^3}\\=6.\left(\frac00\right)\overset{H}{=}6\lim_{x\to\,0}\;\frac{1-\frac{1}{\cos^2x}}{3x^2}=2\cdot\lim_{x\to\,0}\;\frac{\cos^2x-1}{x^2\cdot\cos^2x}=2.\lim_{x\to\,0}\;\frac{-\sin^2x}{x^2.\cos^2x}\\=-2\cdot\lim_{x\to\,0}\;\frac{\sin^2x}{x^2}\cdot\lim_{x\to\,0}\;\frac{1}{\cos^2x}=-2.1.1=-2$