Wiskunde Multiple Choice Vraag

	Welke uitspraak is waar ?
A.	$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}=+\infty\;\;\text{en}\;\;\displaystyle\lim_{x\to 0}\,\frac{1}{1-\sin x-\cos x}=+\infty$
B.	$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}=+\infty\;\;\text{en}\;\;\displaystyle\lim_{x\to 0}\,\frac{1}{1-\sin x-\cos x}=-\infty$
C.	$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}=-\infty\;\;\text{en}\;\;\displaystyle\lim_{x\to 0}\,\frac{1}{1-\sin x-\cos x}=+\infty$
D.	$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}=-\infty\;\;\text{en}\;\;\displaystyle\lim_{x\to 0}\,\frac{1}{1-\sin x-\cos x}=-\infty$
E.	$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}\quad\text{bestaat niet}$

Welke uitspraak is waar ?

$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}=+\infty\;\;\text{en}\;\;\displaystyle\lim_{x\to 0}\,\frac{1}{1-\sin x-\cos x}=+\infty$

$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}=+\infty\;\;\text{en}\;\;\displaystyle\lim_{x\to 0}\,\frac{1}{1-\sin x-\cos x}=-\infty$

$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}=-\infty\;\;\text{en}\;\;\displaystyle\lim_{x\to 0}\,\frac{1}{1-\sin x-\cos x}=+\infty$

$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}=-\infty\;\;\text{en}\;\;\displaystyle\lim_{x\to 0}\,\frac{1}{1-\sin x-\cos x}=-\infty$

$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}\quad\text{bestaat niet}$

	Which statement is correct ?
A.	$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}=+\infty\;\;\text{en}\;\;\displaystyle\lim_{x\to 0}\,\frac{1}{1-\sin x-\cos x}=+\infty$
B.	$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}=+\infty\;\;\text{en}\;\;\displaystyle\lim_{x\to 0}\,\frac{1}{1-\sin x-\cos x}=-\infty$
C.	$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}=-\infty\;\;\text{en}\;\;\displaystyle\lim_{x\to 0}\,\frac{1}{1-\sin x-\cos x}=+\infty$
D.	$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}=-\infty\;\;\text{en}\;\;\displaystyle\lim_{x\to 0}\,\frac{1}{1-\sin x-\cos x}=-\infty$
E.	$\displaystyle\lim_{x\to 0}\,\frac{x}{x-\sin x}\quad\text{does not exist}$