Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\displaystyle \lim_{x \to 0} \: \frac {\cos ax\:-\:1} {x^2}=$	A. \(1\)
B. \(-\frac12a^2\)
C. \(\boldsymbol{- \,\large \frac {8}{a^2}} \)
D. \(\frac12\)
E. \(\frac12 a^2\)

$\displaystyle \lim_{x \to 0} \: \frac {\cos ax\:-\:1} {x^2}=$	A. $1$
	B. $-\frac12a^2$
	C. $\boldsymbol{- \,\large \frac {8}{a^2}} $
	D. $\frac12$
	E. $\frac12 a^2$

1^ste manier :
$\displaystyle\lim_{x\to 0}\:\frac{\cos ax\:-\:1}{x^2}=\displaystyle\lim_{x\to 0}\:\frac{(\cos ax\:-\:1)(\cos ax+1)}{x^2(\cos ax+1)}=\displaystyle\lim_{x\to 0}\:\frac{\cos^2ax\:-\:1}{x^2(\cos ax+1)}\\=\displaystyle\lim_{x\to 0}\:\frac{-\sin^2ax}{x^2(\cos ax+1)}=-a^2\displaystyle\lim_{x\to 0}\:\frac{\sin ax.\sin ax}{ax.ax}.\displaystyle\lim_{x\to 0}\:\frac{1}{\cos ax+1}\\=-a^2.1.1.\frac{1}{1+1}=-\frac{a^2}{2}$
2^de manier : met de regel van de L'Hospital
$\\\displaystyle\lim_{x\to 0}\:\frac{\cos ax\:-\:1}{x^2}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 0}\:\frac{-a\sin ax}{2x}=-\frac{a^2}{2}.\displaystyle\lim_{x\to 0}\:\frac{\sin ax}{ax}=-\frac{a^2}{2}.1=-\frac{a^2}{2}$