Prove the slope of the asymptote of a hyperbola Consider a hyperbola that opens horizontally (left to right):
$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
How do we prove that the slopes of the asymptotes are $\pm \frac{b}{a}$?

Trying to solve it myself:
Rearranging for $y$ I get:
$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
$$a^2(y-k)^2 = b^2(x-h)^2 - a^2b^2$$
$$y = k \pm \sqrt{\frac{b^2(x-h)^2 - a^2b^2}{a^2}}$$
$$y = f(x) = k \pm \frac{b \sqrt{-a^2 + h^2 + x^2 - 2 h x}}{a}$$
So then the slope is
$$\lim_{x\to\infty}\frac{f(x)}{x}$$
or
$$\lim_{x\to\infty}\frac{k}{x} \pm \frac{b \sqrt{-a^2 + h^2 + x^2 - 2 h x}}{ax}$$
$$\lim_{x\to\infty}\frac{k}{x} \pm \frac{b \sqrt{\frac{-a^2 + h^2 + x^2 - 2 h x}{x^2}}}{a}$$
$$\lim_{x\to\infty}\frac{k}{x} \pm \frac{b \sqrt{-\frac{a^2}{x^2} + \frac{h^2}{x^2} + \frac{1}{1} - \frac{2 h}{x}}}{a}$$
$$\lim_{x\to\infty}\frac{k}{\infty} \pm \frac{b \sqrt{-\frac{a^2}{\infty^2} + \frac{h^2}{\infty^2} + \frac{1}{1} - \frac{2 h}{\infty}}}{a}$$
$$\lim_{x\to\infty}0 \pm \frac{b \sqrt{0 + 0 + 1 - 0}}{a}$$
$$\lim_{x\to\infty}\pm \frac{b}{a}$$
The result seems to be right but is my work here correct?
