Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\displaystyle \lim_{x \to +\infty}\: \frac{x+\sqrt{x^2+1}}{x}-\displaystyle \lim_{x \to -\infty}\: \frac{x+\sqrt{x^2+1}}{x}$ is gelijk aan	A. 1
B. 2
C. 0
D. 4
E.

$\displaystyle \lim_{x \to \infty}\: \frac{x+\sqrt{x^2+1}}{x}- \lim_{x \to -\infty}\: \frac{x+\sqrt{x^2+1}}{x}$ is equal to	A. 1
	B. 2
	C. 0
	D. 4
	E. ∞

$\inline\displaystyle\lim_{x\to+\infty}\:\frac{x+\sqrt{x^2+1}}{x}-\lim_{x\to-\infty}\:\frac{x+\sqrt{x^2+1}}{x}\\\lim_{x\to+\infty}\:\frac{x+\sqrt{x^2(1+\frac{1}{x^2})}}{x}-\lim_{x\to-\infty}\:\frac{x+\sqrt{x^2(1+\frac{1}{x^2}})}{x}\\\lim_{x\to\infty}\:\frac{x+x\sqrt{1+\frac{1}{x^2}}}{x}-\lim_{x\to-\infty}\:\frac{x-x\sqrt{\frac{1}{x^2}}}{x}\\\lim_{x\to+\infty}\:\left(1+\sqrt{1+\frac{1}{x^2}}\right)-\lim_{x\to+\infty}\:\left(1-\sqrt{1+\frac{1}{x^2}}\right)=1\sqrt1-(1-\sqrt1)=2&space;$

^{gricha - v6984 - 23.2.13}