Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

A. $\displaystyle\lim_{x\to 0}\,\frac{x}{x-\tan x}=+\infty\quad\text{en}\quad\displaystyle\lim_{x\to 0}\,\frac{1}{x-\cos x}=+\infty$

B. $\displaystyle\lim_{x\to 0}\,\frac{x}{x-\tan x}=+\infty\quad\text{en}\quad\displaystyle\lim_{x\to 0}\,\frac{1}{x-\cos x}=-\infty$

C. $\displaystyle\lim_{x\to 0}\,\frac{x}{x-\tan x}=-\infty\quad\text{en}\quad\displaystyle\lim_{x\to 0}\,\frac{1}{x-\cos x}=+\infty$

D. $\displaystyle\lim_{x\to 0}\,\frac{x}{x-\tan x}=-\infty\quad\text{en}\quad\displaystyle\lim_{x\to 0}\,\frac{1}{x-\cos x}=-\infty$

E. $\displaystyle\lim_{x\to 0}\,\frac{x}{x-\tan x}\quad\text{of}\quad\displaystyle\lim_{x\to 0}\,\frac{1}{x-\cos x}\quad\text{bestaat niet}$

$\displaystyle \lim_{x \to 0}\frac{x}{x-\tan x}\left ( =\frac 00 \right )\overset{H}{=}\displaystyle \lim_{x \to 0}\frac{1}{1-\frac{1}{\cos^2 x}}=\frac{1}{-0}=-\infty$
$\displaystyle \lim_{x \to 0}\frac{1}{1-\cos x}=\frac{1}{+0}=+\infty$

^{gricha - v6986 - 31.5.12}