Considering scaling factor in infinite approximations Trying to arrive at the equations for the asymptotes of the general hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, I rearranged it to $$y=\pm \frac{b}{a}\sqrt{x^2-a^2}$$ considering both positive and negative infinities.
Now, if I neglect the $-a^2$ as x goes to infinity and consider the scaling factor $\frac b a$, I get the correct equation $$y=\pm\frac b a x$$
Why exactly should the scaling factor be considered here? Why is it wrong to just approximate $y=x$, reasoning that x is really large and that therefore the $\frac b a$ shouldn't matter?
 A: In my comment, I tried to illustrate that as $x\to \infty$, it is incorrect to say that $\frac {bx}{a}=x$ by arguing that $\lim_{x\to \infty} \frac {\frac {bx}{a}}{x}=\frac ba \neq 1$.
Of course, the method you have utilized to find equation of asymptote seems rather informal, and I would not have much confidence in it. Rather, I know of two methods to find the required equation, which I list below.
Method 1:
Let the equation of the asymptote be $y=mx+c$. Solving this with equation of hyperbola, we reduce to the quadratic:
$$b^2 x^2 -a^2(mx+c)^2=a^2b^2$$
$$(b^2-a^2m^2)x^2-2a^2mc x-(a^2c^2+a^2b^2)=0$$
Since we do not want any roots, no matter what $a$ or $b$ may be, instead of trying $D<0$ we equate both coefficients of $x$ and $x^2$ to $0$, hence getting $c=0$ and $m=\pm \frac ba$.
Method 2:
We know, for general cubic $$ax^2+2hcy+by^2+2gx+2fy+c=0$$ we have $\Delta=abc+2fgh-bg^2-af^2-ch^2$. To get the required pair of asymptotes, simply choose $c$ such that $\Delta=0$, while changing no other coefficients.
