Wiskunde Multiple Choice Vraag

$\inline\\\displaystyle\lim_{x\to+\infty}\;\left[\sqrt{x^2+3x}-x\right]=$	A. 0
B. 1
C. 1,5
D. 2
E. 3

$\inline\\\displaystyle\lim_{x\to+\infty}\;\left[\sqrt{x^2+3x}-x\right]\\=\lim_{x\to+\infty}\;\frac{(\sqrt{x^2+3x}-x).(\sqrt{x^2+3x}+x)}{\sqrt{x^2+3x}+x}\\=\lim_{x\to+\infty}\;\frac{x^2+3x-x^2}{\sqrt{x^2+3x}+x}=\lim_{x\to+\infty}\;\frac{3x}{x\left(\sqrt{1+\frac3x}+1\right)}\\=\lim_{x\to+\infty}\;\frac{3}{\sqrt{1+\frac3x}+1}=\frac{3}{\sqrt{1+0}+1}=\frac32$

^{gricha - vA167 - 31.3.2026}