Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De limiet $https://latex.codecogs.com/png.image?\large \dpi{110}\displaystyle \lim_{x \to +\infty}\frac{e^2.\, x^2}{e^x}$ is gelijk aan	A. 1
B. 2
C. e²
D. 0
E.

$https://latex.codecogs.com/png.image?\large \dpi{110}\displaystyle \lim_{x \to \infty}\frac{e^2.\, x^2}{e^x}=$	A. 1
B. 2
C. e²
D. 0
E. ∞

1^ste manier : met de regel van del'Hospital
$\\\displaystyle \lim_{x \to +\infty}\frac{e^2.\, x^2}{e^x}\left ( =\frac{+\infty }{+ \infty } \right)\overset{H}{=}\displaystyle \lim_{x \to +\infty}\frac{2e^2x}{e^x}\left (=\frac{+\infty }{+ \infty } \right )\overset{H}{=}\displaystyle \lim_{x \to +\infty}\frac{2e^x}{e^x}=\frac{2e^2}{+\infty}= 0^{(+)}$
2^de manier : met de reeksontwikkeling van e^x
$\\\displaystyle \lim_{x \to +\infty}\frac{e^2.\, x^2}{e^x} =\displaystyle \lim_{x \to +\infty}\frac{e^2x^2}{1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+}=\displaystyle \lim_{x \to +\infty}\frac{e^x}{\frac{1}{x^2}+\frac{1}{x}+\frac{1}{2!}+\frac{x}{3!}+\frac{x^2}{4!}+\cdots }\\=\frac{e^x}{0+0+\frac12+\infty+\infty}=\frac{e^2}{+\infty}=0^{(+)}$