Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\begin{aligned} & \displaystyle \lim_{x \to 0}\:\frac {\ln x - \ln \sin x} {x^2}=\end{aligned}$	A. 2
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$\begin{aligned} &\lim_{x \to 0}\frac{\ln x - \ln \sin x}{x^2} =\lim_{x \to 0}\frac{\ln \frac{x}{\sin x}}{x^2}\left ( =\begin{matrix} 0\\-\\0\end{matrix} \right)\\ &\begin{matrix} H\\ & =\\ {\color{Green} .} \end{matrix} \lim_{x \to 0}\frac{\frac{\sin x}{x}\cdot \frac{\sin x - x.\cos x}{\sin^2x}}{2x} = \lim_{x \to 0}\frac{\sin x}{x}\cdot \lim_{x \to 0}\frac{\sin x - x.\cos x}{2x\sin^2x}\\ & \begin{matrix} H\\ & ={\color{Green} .} \end{matrix} \: 1\cdot \lim_{x \to 0}\frac{\cos x - \cos x + x.\sin x}{2.sin^2x + 4x\sin x\cos x}=\lim_{x \to 0}\frac{x}{2.\sin x + 4\, x\, \cos x}\\ & =\lim_{x \to 0}\frac{1}{2\frac{\sin x}{x}+4.\cos x}=\frac{1}{2.1+4.1}=\frac{1}{6}\end{aligned}$