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$y=\dfrac{2x^2+3x-4}{-4x^2+3x+2}$. 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?

prat
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2 Answers2

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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 $x\to\infty$

$$ \lim_{x\to\infty} y \sim \lim_{x\to\infty} \dfrac{2+\frac3x-\frac4{x^2}}{-4+\frac3x+\frac2{x^2}} = \dfrac{2 + 0 + 0}{-4 + 0 + 0} = -\frac24 = \color{green}{\boxed{-\frac12}} $$

Frank Vel
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  • but the problem is that range is all real – prat Mar 24 '20 at 12:11
  • @prat If you plot the graph (write the formula into wolframalpha, google, or another graphing tool), you will find that part of the function passes through $-1/2$ at $x=2/3$.

    A line being a horizontal asymptote doesn't mean that the function cannot attain the value at any point, but that the function cannot attain the value as $x\to\pm\infty$.

     – Frank Vel Mar 24 '20 at 12:17
  • I've seen that tactic when doing limits. Why multiply by that factor and no other? – Nεo Pλατo Mar 24 '20 at 12:51
  • @Plato $1/x^2$ is the inverse of $x^2$, which is the greatest degree of $x$ in the numerator/denominator. – Frank Vel Mar 24 '20 at 13:51
  • So that's the general method for infinity limits? – Nεo Pλατo Mar 24 '20 at 13:52
  • @Plato for fractions, yes. Note that the term we are multiplying is identical to $1$, as long as $x\neq0$. $\frac{\frac1x}{\frac1x} = 1$, $x\neq0$. – Frank Vel Mar 24 '20 at 13:54
  • There was also another one. If it's not too much how would you start solving this one: $\displaystyle \lim_{x \to 0} (\sin x)^{1-\cos{x}}$ – Nεo Pλατo Mar 24 '20 at 13:57
  • @plato I don't think comments are appropriate for solving questions, and that one would be beyond the scope of this one. I suggest you open a new question. Good luck! – Frank Vel Mar 24 '20 at 15:39
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I believe the question is a bit ill-stated, you must be wanting to find a horizontal asymptote on a specific branch of the above polynomial. Plotting the above function in Desmos, we obtain this picture.

This clearly shows that if you took into picture the different branches, the range of the above would in fact be $(- \infty, \infty)$. However, if you focus on a single branch, the asymptote would infact be $\frac{-1}{2}$ as obtained.

Shikhar Jaiswal
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  • ok, so the problem is in the definition of horizontal asymptotes, so what should be the definition of horizontal asymptotes – prat Mar 24 '20 at 12:21
  • @prat The same as mentioned by Frank Vel, horizontal asymptotes can be only found at extremes. The difference is in perspective, you can apply the second method as you mentioned, but only on specific branches, not the whole polynomial. – Shikhar Jaiswal Mar 24 '20 at 12:25