Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

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

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to -\infty} \: (\sqrt{x^2+x} +x)$ is equal to	A.
	B. −
	C. 0
	D.
	E. ∞

$\begin{aligned} \displaystyle\lim_{x\to-\infty}\left(\sqrt{x^2+x}+x\right)=\displaystyle\lim_{x\to-\infty}\frac{\left(\sqrt{x^2+x}+x\right)\left(\sqrt{x^2+x}-x\right)}{\left(\sqrt{x^2+x}-x\right)}\;\\=\displaystyle\lim_{x\to-\infty}\frac{x^2+x-x^2}{\sqrt{x^2+x}-x}=\displaystyle\lim_{x\to-\infty}\frac{x}{\sqrt{x^2+x}-x}=\displaystyle\lim_{x\to-\infty}\frac{x}{\sqrt{x^2\left(1+\frac{1}{x}\right)}-x}\;\\=\displaystyle\lim_{x\to-\infty}\frac{x}{-x\sqrt{1+\frac{1}{x}}-x}=\displaystyle\lim_{x\to-\infty}\frac{1}{-\sqrt{1+\frac{1}{x}}-1}=\frac{1}{-1-1}=-\frac12 \end{aligned}$