Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De limiet $https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{\underset{<}{x \to 4}} \: \frac {\sin x - \sin 4} {(x\,-\,4)^2}$ is gelijk aan	A. 1
B.
C.
D.
E.

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{{x \to 4^-}} \: \frac {\sin x - \sin 4} {(x\,-\,4)^2}$ is equal to	A. 1
	B.
	C.
	D.
	E. ∞

1^ste manier : zonder de regel van de l'Hospital
$\\\displaystyle\lim_{{x\to 4^-}}\:\frac{\sin x-\sin 4}{(x\,-\,4)^2}=\displaystyle\lim_{{x\to 4^-}}\:\frac{2\cos\frac{x+4}{2}\sin\frac{x-4}{2}}{(x-4)^2}=\displaystyle\lim_{{x\to 4^-}}\:\cos\frac{x+4}{2}.\displaystyle\lim_{{x\to 4^-}}\:\frac{\sin\frac{x-4}{2}}{\frac{x-4}{2}}.\displaystyle\lim_{{x\to 4^-}}\:\frac{1}{x-4}\\=\cos 4.1.\frac{1}{-0}=+\infty$
2^de manier : metde regel van de l'Hospital
$\\\displaystyle\lim_{{x\to 4^-}}\:\frac{\sin x-\sin 4}{(x\,-\,4)^2}\left(=\frac00\right)\overset{H}={}\displaystyle\lim_{{x\to 4^-}}\:\frac{\cos x}{2(x-4)}=\frac{neg.getal}{-0}=+\infty$

^{gricha - v5563 - 19.11.13}