Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\large \dpi{110}\displaystyle\: \lim_{x \to 16}\: \frac{x-16}{{\sqrt[4]{x}-2}$ is gelijk aan	A. 2¹
B. 2²
C. 2³
D. 2⁴
E. 2⁵

$https://latex.codecogs.com/png.image?\large \dpi{110}\displaystyle\: \lim_{x \to 16}\: \frac{x-16}{{\sqrt[4]{x}-2}$ is equal to	A. 2¹
B. 2²
C. 2³
D. 2⁴
E. 2⁵

1^ste manier : m.b.v. toegevoegde vormen
$\\\lim_{x\to16}\frac{x-16}{\sqrt[4]{x}-2}=\lim_{x\to16}\frac{\left(x-16\right)\left(\sqrt[4]{x}+2\right)}{\left(\sqrt[4]{x}-2\right)\left(\sqrt[4]{x}+2\right)}=\lim_{x\to16}\frac{\left(x-16\right)\left(\sqrt[4]{x}+2\right)}{\left(\sqrt x-4\right)}\\=\lim_{x\to16}\frac{\left(x-16\right)\left(\sqrt[4]{x}+2\right)\left(\sqrt x+4\right)}{\left(\sqrt x-4\right)\left(\sqrt x+4\right)}=\lim_{x\to16}\;\left(\sqrt[4]{x}+2\right)\left(\sqrt x+4\right)\\=\left(2+2\right)\left(4+4\right)=4.8=32$
2^de manier : m.b.v. de regel van de l'Hospital
$\\\lim_{x\to16}\frac{x-16}{\sqrt[4]{x}-2}\lim_{x\to16}\frac{x-16}{x^\frac{1}{4}-2}\left(=\frac{0}{0}\right)\buildrel H\over=\lim_{x\to16}\;\frac{1}{\frac{1}{4}x^{\frac{1}{4}-1}}=\lim_{x\to16}\;\frac{1}{\frac{1}{4}x^{-\frac{3}{4}}}\\=\lim_{x\to16}\;.x^\frac{3}{4}=4.16^\frac{3}{4}=4.(2^4)^\frac{3}{4}=4.2^3=4.8=32$