Where does the equation of asymptotes of a hyperbola come from? It's known that the asymptotes of a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is given by $y=\pm\frac{b}{a}x$ if $a>b$.
I  tried to find a proof of the fact that why the equations of these asymptotes are like that,however the only reference (Thomas calculus  book) that I found explained 
that the two asymptotes are derived by letting $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0$.

It would be highly appreciated if someone prove why the equation of the asymptotes have such form.

Edited to do it properly -- see below
Original post:
We have
$$y=b\sqrt{\frac{x^2}{a^2}-1}=\frac{b}{a}x\sqrt{1-\frac{a^2}{x^2}}$$
And as $x\to\pm\infty$, $\sqrt{1-\frac{a^2}{x^2}}\to 1$.
End of original post
But as mentioned in the comments, it is not enough to show that $\frac{y}{bx/a}\to 1$. We have to show that $y-\frac{b}{a} x\to 0$:
$$y-\frac{b}{a}x=\frac{b}{a}x\left(\sqrt{1-\frac{a^2}{x^2}}-1\right)$$
But
$$1-\frac{a^2}{x^2}\le\sqrt{1-\frac{a^2}{x^2}}<1$$
So
$$\left|\sqrt{1-\frac{a^2}{x^2}}-1\right|\le\frac{a^2}{x^2}$$
Therefore
$$\left|y-\frac{b}{a}x\right|\le\frac{b}{a}|x|\cdot\frac{a^2}{x^2}=\frac{ba}{|x|}$$
which tends to $0$ as $x\to\pm\infty$.
