Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{{x \to 0}}\; (\sin x + \cos x)^{\frac{1}{2x}}$ is gelijk aan	A. 1
B.
C. e²
D. ${\color{DarkBlue} \frac{1}{e^2}}$
E.

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{{x \to 0}}\; (\sin x + \cos x)^{\frac{1}{2x}}$ is equal to	A. 1
	B.
	C. e²
	D. ${\color{DarkBlue} \frac{1}{e^2}}$
	E.

$\displaystyle\lim_{x\to 0}\;(\sin{x}+\cos{x})^\frac{1}{2x}\left(=1^\infty\right)\overset{\;x=e^{\ln x}}{=\!=}\displaystyle\lim_{x\to 0}\;e^{\ln{(}\sin{x}+\cos{x})^\frac{1}{2x}}=\displaystyle\lim_{x\to 0}\;e^\frac{ln{(}\sin{x}+\cos{x})}{2x}$
$EXPO=\displaystyle\lim_{x\to 0}\;\frac{\ln\,(\sin x+\cos x)}{2x}$
$\displaystyle\lim_{x\to 0}\;EXPO=\displaystyle\lim_{x\to 0}\;\frac{\ln\,(\sin x+\cos x)}{2x}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 0}\;\frac{\frac{\cos x-\sin x}{\sin x+\cos x}}{2}=\frac{\frac11}{1}=\frac12$
$\displaystyle\lim_{x\to 0}\;(\sin{x}+\cos{x})^{\frac{1}{2x}}=e^{\frac12}=\sqrt e\approx 1,6487$