Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\large \dpi{110}\displaystyle\: \lim_{x \to +\infty}\: x^{-\frac 2x}$ is gelijk aan	A. 1
B. 2
C. 0
D. e⁻²
E.

$\displaystyle\: \lim_{x \to \infty}\: x^{-\frac 2x}$ is equal to	A. 1
	B. 2
	C. 0
	D. e⁻²
	E. ∞

$\\\displaystyle \lim_{x \to +\infty} x{^\frac 2x} \left ( =(+\infty)^0 \right ) =\displaystyle \lim_{x \to +\infty} e^{\ln x^{-\frac 2x}}\\=\displaystyle \lim_{x \to +\infty} e^{-\frac 2x .\ln x} =\displaystyle \lim_{x \to +\infty} e^{-\frac{2.\ln x}{x}}\\=e^{-\displaystyle \lim_{x \to +\infty}\frac{2.\ln x}{x} }=e^{-\displaystyle \lim_{x \to +\infty} \frac {\frac 2x}{1}}\\=e^{-\displaystyle \lim_{x \to +\infty}\frac 2x }=e^0 = 1$
In één van de stappen hebben we de
regel van de l'Hospital toegepast.