Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De limiet van $\frac{\sin x}{1-\cos x}$ is als	A. $x\:{\underset{>}{\to}0}$
	B. $x\:{\underset{>}{\to}\frac{\pi}{2}}$
	C. $x\:{\underset{>}{\to}\pi}$
	D. $x\:{\underset{<}{\to}\frac{\pi}{2}}$
	E. $x\:{\underset{<}{\to}0}$
	F. $x\to+\infty$

[ 5-6982 - op net sinds 18.9.2024-(E)- ]

Translation in E N G L I S H

IN CONSTRUCTION

Oplossing - Solution

$\displaystyle\lim_{\underset{>}{x\to 0}}\frac{\sin x}{1-\cos x}=\displaystyle\lim_{\underset{>}{x\to 0}}\frac{\sin x(1+\cos x)}{1-\cos^2x}=\displaystyle\lim_{\underset{>}{x\to 0}}\frac{\sin x(1+\cos x)}{\sin^2x}=\displaystyle\lim_{\underset{>}{x\to 0}}\frac{1+\cos x}{\sin x}=\frac{1+1}{+0}=+\infty$