Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{{x \to 0}}\; \frac {\tan x \: - \: \sin x} {x^3}$ is gelijk aan	A. 1
B.
C.
D. 0
E.

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{{x \to 0}}\; \frac {\tan x \: - \: \sin x} {x^3} =$	A. 1
B.
C.
D. 0
E. ∞

1^ste manier : zonder de regel van de l'Hospital
$\inline\displaystyle\lim_{{x\to 0}}\;\frac{\tan x\:-\:\sin x}{x^3}=\lim_{{x\to 0}}\;\left(\frac{\sin x}{x}.\frac{\frac{1}{\cos x}-1}{x^2}\right)=\lim_{{x\to 0}}\;\left(\frac{\sin x}{x}.\frac{1-\cos x}{x^2.\cos x}\right)\\=\lim_{{x\to 0}}\;\left(\frac{\sin x}{x}.\frac{(1-\cos x)(1+\cos x)}{x^2.\cos x.(1+\cos x)}\right)=\lim_{{x\to 0}}\;\left(\frac{\sin x}{x}.\frac{\sin^2x}{x^2.\cos x.(1+\cos x)}.\right)\\=\lim_{{x\to 0}}\;\left(\frac{\sin x}{x}.\frac{\sin^2x}{x^2}.\frac{1}{\cos x.(1+\cos x)}\right)=1.1.\frac{1}{1.2}=\frac12$
2^de manier : : met de regel van de l'Hospital
$\inline\\\displaystyle\lim_{{x\to 0}}\;\frac{\tan x\:-\:\sin x}{x^3}\left(=\frac00\right)\buildrel H\over=\displaystyle\lim_{{x\to 0}}\;\frac{\frac{1}{\cos^2x}-\cos x}{3x^2}\buildrel H\over=\lim_{{x\to 0}}\;\frac{-\frac{1.2.\cos x(-\sin x)}{\cos^4x}+\sin x}{6x}\\=\lim_{{x\to 0}}\;\frac{\frac{1.2.\sin x}{\cos^3x}+\sin x}{6x}=\lim_{{x\to 0}}\;\left[\frac{\sin x}{6x}.\left(\frac{2}{\cos^3x}+1\right)\right]=\frac16(2+1)=\frac12$

^{gricha - v6639 - 5.10.2017}