Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to +\infty}\sqrt[n]{\frac{n^2}{3^n+5^n}}$	A.
B.
C.
D. e⁻³
E. e^− 5

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to \infty}\sqrt[n]{\frac{n^2}{3^n+5^n}}=$	A.
	B.
	C.
	D. e⁻³
	E. e^− 5

$\lim_{n\rightarrow+\infty}\;\sqrt[n]{\frac{n^2}{3^n+5^n}}=\lim_{n\rightarrow+\infty}\;\frac{n^\frac{2}{n}}{\sqrt[n]{3^n+5^n}}\\=\lim_{n\rightarrow+\infty}\;\frac{e^{ln{\;}\;n^\frac{2}{n}}}{\sqrt[n]{5^n.\left(\frac{3^n}{5^n}+1\right)}}=\lim_{n\rightarrow+\infty}\;\frac{e^{\frac{2}{n}.ln{\;}n}}{\sqrt[n]{5^n.\left(\frac{3^n}{5^n}+1\right)}}\;\left(\ast\right)$
$\text{en vermits}\;\lim_{n\rightarrow+\infty}\;\frac{ln{\;}n}{n}\left(=\frac{+\infty}{+\infty}\right)\buildrel H\over=\lim_{n\rightarrow+\infty}\;\frac{\frac{1}{n}}{1}=\lim_{n\rightarrow+\infty}\frac{1}{n}=0$
$\text{is\;(*)\;gelijk aan}\;\;\frac{e^{2.0}}{5.\sqrt[5]{0+1}}=\frac{e^0}{5}=\frac{1}{5}$