Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0}\:\frac {x} {D\frac{x}{\sin x}} =$ De noemer is de afgeleide van ^x/_{sin x} )	A. − 1
B. 0
C. 1
D. 2
E. 3

$\displaystyle \lim_{x \to 0}\:\frac {x} {D\frac{x}{\sin x}} =$ _{( D = Derivative )}	A. − 1
	B. 0
	C. 1
	D. 2
	E. 3

$\begin{aligned}&\displaystyle\lim_{x\to 0}\:\frac{x}{D\frac{x}{\sin x}}=\displaystyle\lim_{x\to 0}\:\frac{x}{\frac{\sin x-x.\cos x}{\sin^2x}}=\displaystyle\lim_{x\to 0}\:\frac{x.\sin^2x}{\sin x-x.\cos x}\left(\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 0}\:\frac{\sin^2x+x.2.\sin x.\cos x}{\cos x-\cos x+x\sin x}\\&=\displaystyle\lim_{x\to 0}\:\frac{\sin^2x+x.\sin x.2.\cos x}{x.\sin x}=\displaystyle\lim_{x\to 0}\:\frac{\sin x}{x}+\displaystyle\lim_{x\to 0}\:2\,\cos x=1+2=3\end{aligned}$