Finding the vertical asymptote of $y = \pm2\sqrt\frac{x}{x-2}$ Find the value of vertical asymptotes.
$$ y = \pm2\sqrt\frac{x}{x-2}$$
There are two fuctions right? so,
$$ y = f_1(x)= +2\sqrt\frac{x}{x-2}$$
$$ y = f_2(x)= -2\sqrt\frac{x}{x-2}$$
I'm good until here.
Now, the book says
$$\lim_{x\to2^+}f_1(x) = \lim_{x\to2^+} 2\sqrt\frac{x}{x-2} = + \infty$$
The book is a little old. Some parts were ripped.
Could anyone please tell me why we say that the limit approaches 2 from the right side??

The book most probably deals with real-valued functions. So you can't take the limit from the left since as $x$ approaches $2$ from the left, the quantity $\tfrac x{x-2}$ is still negative, and so you can't talk about $2\sqrt{\tfrac x{x-2}}$ because that's not a defined quantity.
