Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\displaystyle \lim_{x \to 0}\, \frac{x^2}{1-\sec 2x}=$	A. 1
B. 2
C. − 1
D.
E.

$\displaystyle \lim_{x \to 0}\, \frac{x^2}{1-\sec 2x}=$	A. 1
B. 2
C. − 1
D.
E.

$\\\displaystyle\lim_{x\to 0}\,\frac{x^2}{1-\sec 2x}=\displaystyle\lim_{x\to 0}\,\frac{x^2.\cos2x}{\cos^2x-1}=\displaystyle\lim_{x\to 0}\,\frac{x^2.\cos2x}{-2\sin^2x}\\=\displaystyle\lim_{x\to 0}\,\frac{x^2}{\sin^2x}.\displaystyle\lim_{x\to 0}\,\frac{\cos 2x}{-2}=1.\frac{1}{-2}=-\frac12$
$\\\displaystyle\lim_{x\to 0}\,\frac{x^2}{1-\sec 2x}=\displaystyle\lim_{x\to 0}\,\frac{x^2.\cos2x}{\cos^2x-1}=\displaystyle\lim_{x\to 0}\,\frac{x^2.\cos2x}{-2\sin^2x}\\=\displaystyle\lim_{x\to 0}\,\frac{x^2}{\sin^2x}.\displaystyle\lim_{x\to 0}\,\frac{\cos 2x}{-2}=1.\frac{1}{-2}=-\frac12$