Asymptotes of hyperbola by considering x tends to infinity I saw this derivation of equations of the asymptotes of hyperbola and it goes like this...
For a standard hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, rearranging the terms we get
$y=\pm\frac{b}{a}x\sqrt{1-\frac{a^2}{x^2}}$
So as $x\rightarrow\infty$, $\frac{a^2}{x^2}\rightarrow 0$ and hence $y\rightarrow\pm\frac{b}{a}x$.
Therefore the equations of the asymptotes are $y=\pm\frac{b}{a}x$.
I am not very sure if this derivation is correct. Would appreciate if you can share your opinion. Like if there is any step above that is not true in general. 

I think you are right to be suspicious of this method.
Let's apply it to the hypberbola given by the equation
$$ \frac{x^2}{2^2} + x -\frac{y^2}{2^2}=1 .$$
Solving for $y$ as in the method in the question, we get
$$ y = \pm x \sqrt{1 + \frac 4x - \frac 4{x^2}},$$
and as $x\to\infty,$ we find that $\sqrt{1 + \frac 4x - \frac 4{x^2}}\to 1,$
so this method derives the asymptotes $y = x$ and $y = -x.$
But the actual asymptotes are $y = 2 + x$ and $y = 2-x.$
The pitfall in this method can be seen by applying it to any straight line.
For the equation $y = mx + b,$ We factor $x$ out of the right side to obtain
$$ y = x \left(m + \frac bx\right), $$
and then $\left(m + \frac bx\right)\to m$ as $x\to\infty,$
so the method yields $y = mx.$
What the method is actually finding is the directions from the origin to the points at infinity on the curve, which gives the slopes of the asymptotes but not the $x$- or $y$-intercepts.
It will coincidentally give the correct result when the asymptote happens to pass through the origin, as we can predict will happen with
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ due to symmetry,
but that's an extra fact that has to be shown and it works only in that special case.
However, the method does find the slope of each asymptote.
We can then find the $y$-intercept by taking the difference between the curve and a line through the origin with the same slope as the asymptote.
Taking the hyperbola $ \frac{x^2}{2^2} + x -\frac{y^2}{2^2}=1$ again,
taking the solution $y = \sqrt{x^2 + 4x -4}$ and
comparing it with the line $y = x,$
we find that
$$ \lim_{x\to\infty} (y_\mathrm{\,hyperbola} - y_\mathrm{\,line})
= \lim_{x\to\infty} \sqrt{x^2 + 4x -4} - x = 2.$$
Hence the difference between the curve $y = \sqrt{x^2 + 4x -4}$ 
and the line $y = 2 + x$ as $x\to\infty$ is zero.
This correctly predicts that $y = 2 + x$ is an asymptote.
Aretino's answer shows how this method is correctly applied to the hyperbola in the question.
