Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to +\infty}\: \frac{x+\sqrt{x^2+1}}{x}-\displaystyle \lim_{x \to -\infty}\: \frac{x+\sqrt{x^2+1}}{x}$ is gelijk aan	A. 1
B. 2
C. 0
D. 4
E.

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to \infty}\: \frac{x+\sqrt{x^2+1}}{x}-\displaystyle \lim_{x \to -\infty}\: \frac{x+\sqrt{x^2+1}}{x}$ is equal to	A. 1
	B. 2
	C. 0
	D. 4
	E. ∞

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