Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De limiet $https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: \frac {x+\tan x} {1+x-\cos x}$ is gelijk aan	A. 1
B. 2
C.
D.
E.

The limit $https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: \frac {x+\tan x} {1+x-\cos x}$ is equal to	A. 1
B. 2
C.
D.
E.

1^ste manier : zonder de regel van de L'Hospital
$\\\displaystyle\lim_{x\to 0}\:\frac{x+\tan x}{1+x-\cos x}=\displaystyle\lim_{x\to 0}\:\frac{1+\frac{\tan x}{x}}{\frac1x+1-\frac{\cos x}{x}}=\frac{\displaystyle\lim_{x\to 0}\left(1+\frac{\tan x}{x}\right)\:}{1+\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x}\:}=\frac{1+1}{1+\displaystyle\lim_{x\to 0}\:\frac{1-\cos^2x}{x(1+\cos x)}}\\=\frac{2}{1+\displaystyle\lim_{x\to 0}\:\frac{\sin^2x}{x(1+\cos x)}}=\frac{2}{1+\displaystyle\lim_{x\to 0}\frac{\sin x}{x}.\displaystyle\lim_{x\to 0}\:\frac{\sin x}{1+\cos x}}=\frac{2}{1+1.\frac02}=2$
2^de manier : met de regel van de L'Hospital
$\\\displaystyle\lim_{x\to 0}\:\frac{x+\tan x}{1+x-\cos x}\left(\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 0}\:\frac{1+\frac{1}{\cos^2x}}{0+1+\sin x}=\frac{1+1}{0+1+0}=2$