Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De limiet $\lim_{x \to 0}\,\frac{1-\cos 2x}{x^2}$ is	A. gelijk aan	$\lim_{x \to 0}\,\frac{1-\cos x}{x^2}$
B. twee keer
C. de helft van
D. vier keer
E. acht keer

The limit $\lim_{x \to 0}\,\frac{1-\cos 2x}{x^2}$ is	A. equal to	$\lim_{x \to 0}\,\frac{1-\cos x}{x^2}$
B. two times
C. half of
D. four times
E. eight times

1^ste manier :
$\displaystyle\lim_{x\to 0}\frac{1-\cos 2x}{x^2}=\displaystyle\lim_{x\to 0}\frac{2\sin^2x}{x^2}=2\displaystyle\lim_{x\to 0}\frac{\sin x}{x}\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=2.1.1=2$
$\displaystyle\lim_{x\to 0}\frac{1-\cos 2x}{x^2}=\displaystyle\lim_{x\to 0}\frac{2\sin^2\frac x2}{x^2}=2.\frac14\displaystyle\lim_{x\to 0}\frac{\sin\frac x2}{\frac x2}\displaystyle\lim_{x\to 0}\frac{\sin\frac x2}{\frac x2}=\frac12.1.1=\frac12$
Je dat de eerste limiet vier keer groter is dan de tweede.
2^de manier : met de regel van de l'Hospital
$\displaystyle\lim_{x\to 0}\frac{1-\cos 2x}{x^2}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 0}\frac{0+2\sin 2x}{2x}=2\displaystyle\lim_{x\to 0}\frac{\sin 2x}{2x}=2$
$\displaystyle\lim_{x\to 0}\frac{1-\cos 2x}{x^2}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 0}\frac{0+\sin x}{2x}=\frac12\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=\frac12$
Je dat de eerste limiet vier keer groter is dan de tweede.
3^de manier :
$\\\frac{\displaystyle\lim_{x\to 0}\frac{1-\cos 2x}{x^2}}{\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x^2}}=\displaystyle\lim_{x\to 0}\frac{\frac{1-\cos 2x}{x^2}}{\frac{1-\cos x}{x^2}}=\displaystyle\lim_{x\to 0}\frac{1-\cos2x}{1-\cos x}\\=\displaystyle\lim_{x\to 0}\frac{2\sin^2x}{2\sin^2\frac x2}=\displaystyle\lim_{x\to 0}\frac{\sin x}{x}.\frac{\sin x}{x}.\frac{\frac x2}{\sin\frac x2}.\frac{\frac x2}{\sin\frac x2}.4=1.1.1.1.4=4$
$\\\frac{\displaystyle\lim_{x\to 0}\frac{1-\cos 2x}{x^2}}{\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x^2}}=\displaystyle\lim_{x\to 0}\frac{\frac{1-\cos 2x}{x^2}}{\frac{1-\cos x}{x^2}}=\displaystyle\lim_{x\to 0}\frac{1-\cos2x}{1-\cos x}\\=\displaystyle\lim_{x\to 0}\frac{2\sin^2x}{2\sin^2\frac x2}=\displaystyle\lim_{x\to 0}\frac{\sin x}{x}.\frac{\sin x}{x}.\frac{\frac x2}{\sin\frac x2}.\frac{\frac x2}{\sin\frac x2}.4=1.1.1.1.4=4$
4^de manier :
$\\\frac{\displaystyle\lim_{x\to 0}\frac{1-\cos 2x}{x^2}}{\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x^2}}=\displaystyle\lim_{x\to 0}\frac{\frac{1-\cos 2x}{x^2}}{\frac{1-\cos x}{x^2}}=\displaystyle\lim_{x\to 0}\frac{1-\cos2x}{1-\cos x}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 0}\frac{0+2\sin2x}{0+\sin x}\left(=\frac00\right)\\\overset{H}{=}\displaystyle\lim_{x\to 0}\frac{4\cos2x}{\cos x}=\frac{4.\cos0}{\cos0}=4$
$\\\frac{\displaystyle\lim_{x\to 0}\frac{1-\cos 2x}{x^2}}{\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x^2}}=\displaystyle\lim_{x\to 0}\frac{\frac{1-\cos 2x}{x^2}}{\frac{1-\cos x}{x^2}}=\displaystyle\lim_{x\to 0}\frac{1-\cos2x}{1-\cos x}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 0}\frac{0+2\sin2x}{0+\sin x}\left(=\frac00\right)\\\overset{H}{=}\displaystyle\lim_{x\to 0}\frac{4\cos2x}{\cos x}=\frac{4.\cos0}{\cos0}=4$