Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\displaystyle\: \lim_{x \to 0}\: \frac{1-\cos 3x}{1-\frac{\sin x}{x}}$ is gelijk aan	A. 3⁰
B. 3¹
C. 3²
D. 3³
E. 3⁴

$\displaystyle\: \lim_{x \to 0}\: \frac{1-\cos 3x}{1-\frac{\sin x}{x}}$ is equal to	A. 3⁰
	B. 3¹
	C. 3²
	D. 3³
	E. 3⁴

$\text{We steunen op\;:\;}\displaystyle\lim_{x\to0}\frac{\sin ax}{bx}=\displaystyle\lim_{x\to0}\frac{ax}{\sin bx}=\frac ab$
$\begin{aligned}\ lim_{x\to 0}\:\frac{1-\cos 3x}{1-\frac{\sin x}{x}}= \lim_{x\to 0}\:\frac{x-x.\sin 3x}{x-\sin x}\left(=\frac00\right)\overset{H}{=} \lim_{x\to 0}\:\frac{1-\cos 3x+3x\sin x}{1-\cos x}\left(=\frac00\right)\\\overset{H}{=} \lim_{x\to 0}\:\frac{0+3\sin 3x+3\sin 3x+9x\cos 3x}{\sin x}= 3\lim_{x\to 0}\:\left(\frac{\sin3x}{x}+\frac{\sin3x}{x}+\frac{\sin3x}{x}.\cos x\right)\\ =3(3+3+3.1)=3.9=27 \end{aligned}$