Where does the equation of asymptotes of a hyperbola come from?

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 $x\to\pm\infty$.

The other answers have tried to give more rigorous arguments, which I would like to complement with a heuristic way which doesn't need further manipulation of the equation:

From looking at the hyperbola, it is obvious that the asymptotes are lines that the curve approaches when $x$ and $y$ become very large, in particular larger than $a$ or $b$ (BTW, the condition $a<b$ seems unnecessary to me). Then, in the defining equation$$\frac{x²}{a²}-\frac{y²}{b²}=1\,,$$ you have two large numbers on the left-hand side whose difference is $1$. In other words, their difference is much smaller than the numbers themselves, and it becomes a good approximation to just neglect the $1$ on the right-hand side. Furthermore, the approximation becomes better which increasing $x$ and $y$. Thus, $$\frac{x²}{a²}-\frac{y²}{b²}=0\,,$$ is at least a good candidate for the equation of the asymptotes.