Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\large \dpi{110}\displaystyle\: \lim_{x \to 2}\: \frac{2^{x+2}-16}{4^x -16}$ is gelijk aan	A. 1
B. 2
C. een negatief getal
D. 4
E.

$https://latex.codecogs.com/png.image?\large \dpi{110}\displaystyle\: \lim_{x \to 2}\: \frac{2^{x+2}-16}{4^x -16}$ is equal to	A. 1
	B. 2
	C. a negative number
	D. 4
	E.

1^ste manier : zonder de regel van de l'Hospital
$https://latex.codecogs.com/png.image?\dpi{110}\\\displaystyle\: \lim_{x \to 2}\: \frac{2^{x+2}-16}{4^x -16}=\displaystyle\: \lim_{x \to 2}\:\frac{4.2^x-4.4}{(2^2)^x-16} =\displaystyle\: \lim_{x \to 2}\:\frac{4.(2^x-4)}{(2^2)^x-16} =\displaystyle\: \lim_{x \to 2}\:\frac{4.(2^x-4)}{(2^x-4)(2^x+4)} \\{\color{Green} .}\qquad\qquad\qquad=\displaystyle\: \lim_{x \to 2}\:\frac{4}{2^x+4} =\displaystyle\: \lim_{x \to 2}\:\frac{4}{4+4}=\cdots$
2^de manier : met de regel van de l'Hospital
$https://latex.codecogs.com/png.image?\dpi{110}\\\displaystyle\: \lim_{x \to 2}\: \frac{2^{x+2}-16}{4^x -16}\left ( =\frac00 \right )\overset{H}{=}\displaystyle\: \lim_{x \to 2}\: \frac{2^{x+2}.\ln 2}{4^x.\ln 4}=\frac{2^4.\ln2}{4^2.\ln4}=\frac{\ln2}{\ln4}=\frac{\ln2}{2.\ln2}=\cdots$