<s>[INST] $y=\dfrac{2x^2+3x-4}{-4x^2+3x+2}$. Find its horizontal asymptotes? $y=\dfrac{2x^2+3x-4}{-4x^2+3x+2}$. Find its horizontal asymptotes?
My attempt is as follows:-
Horizontal asymptotes are the horizontal lines which signify the values of $y$ which graph cannot ever attain. 
There are two ways to find horizontal asymptotes for the rational function:
$1)$ If degree of numerator is equal to degree of denominator, then $y=\dfrac{\text {leading coefficient of numerator }}{\text{leading coefficient of denominator}}$
So here we have degree of denominator and numerator as equal, so $y=-\dfrac{1}{2}$ will be the horizontal asymptote.
It means graph cannot ever touch $y=-\dfrac{1}{2}$
$2)$ Another way is by finding range, elements which are not in the range will correspond to horizontal asymptotes.
Let's find out the range, 
$$-4x^2y+3xy+2y=2x^2+3x-4$$
$$x^2(2+4y)+3x(1-y)-4-2y=0$$
$$D\ge0$$
$$9(1+y^2-2y)+4(4+2y)(2+4y)\ge0$$
$$9+9y^2-18y+4(8+20y+8y^2)\ge0$$
$$41y^2+62y+41\ge0$$
This is always greater than equal to zero because $D=62^2-4\cdot1681<0$
So this indicates that the range is $\left(-\infty,\infty\right)$. So it means there should be no horizontal asymptotes, but by the first way we got $y=-\dfrac{1}{2}$ as horizontal aysmptote.
What am I missing here?
 [/INST] Another way of finding the horizontal asymptotes is by multiplying by $\frac{\frac1{x^2}}{\frac1{x^2}}$
$$ 
y = \dfrac{2x^2+3x-4}{-4x^2+3x+2} \sim \dfrac{2x^2+3x-4}{-4x^2+3x+2}\frac{\frac1{x^2}}{\frac1{x^2}} = \dfrac{2+\frac3x-\frac4{x^2}}{-4+\frac3x+\frac2{x^2}}
$$
and find the limit as $x\to\infty$
$$
\lim_{x\to\infty} y \sim \lim_{x\to\infty} \dfrac{2+\frac3x-\frac4{x^2}}{-4+\frac3x+\frac2{x^2}} = \dfrac{2 + 0 + 0}{-4 + 0 + 0} = -\frac24 = \color{green}{\boxed{-\frac12}}
$$
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