Understanding horizontal asymptotes of non-even rational functions


by lordofpi
Tags: asymptote, horizontal, infinity, limits, rational
lordofpi
lordofpi is offline
#1
Dec12-12, 02:29 PM
P: 17
I think the technique that is to be used for these types of problems, but I just am having trouble grasping why it is permitted. I have no problem with any homework, but it just doesn't seem right. Maybe my text is just not being clear (Larson, 9th, btw)

Given a function [tex]\frac{ax}{\sqrt{bx^2}}[/tex], where [itex]a[/itex] and [itex]b[/itex] are constants, in order to find the horizontal asymptotes, we must take the limit the function as [itex]x[/itex] approaches [itex]\infty[/itex] and [itex]-\infty[/itex].

For [itex]\infty[/itex], we proceed by multiplying the expression by the multiplicative identity 1, which equals [itex]\frac{1}{1}=\frac{\frac{1}{x}}{\frac{1}{x}}[/itex] Since [itex]\frac{1}{x} = \frac{1}{\sqrt{x^2}}[/itex], we can use this second expression in the denominator of our multiplier, which can now be expressed as [itex]\frac{\frac{1}{x}}{\sqrt{\frac{1}{x^2}}}[/itex].

[tex]\begin{align*}\frac{ax}{\sqrt{bx^2}} \cdot \frac{\frac{1}{x}}{\sqrt{\frac{1}{x^2}}} &= \frac{\frac{ax}{x}} {\frac{\sqrt{bx^2}}{\sqrt{x^2}}} = \frac{\frac{ax}{x}}{\sqrt{\frac{bx^2}{x^2}}} = \frac{a}{\sqrt{b}}
\end{align*}[/tex]

This all makes sense to me. However, why, when seeking the asymptote at [itex]-\infty[/itex] are we allowed to substitute [itex]\frac{1}{x}=-\sqrt{\frac{1}{x^2}}[/itex] for our multiplier. Or at least this is what my textbook does. My guess is that there is very logical explanation, and it certainly makes sense to me in the sense that it solves the problem, but I just can't see why this is permitted. Thanks all.
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Michael Redei
Michael Redei is offline
#2
Dec12-12, 03:07 PM
P: 181
Simple reason: when ##x## is negative, you have ##\sqrt{x^2}=-x##.

In general, ##\sqrt{x^2}=|x|##, and not simply ##x##.
lordofpi
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#3
Dec12-12, 11:58 PM
P: 17
Thanks for the reply Michael. I guess where I am confused is that we are using [itex]\frac{1}{x}[/itex] in the numerator of our multiplier, but then in the denominator we are using [itex]\frac{1}{-\sqrt{x^2}}[/itex]. We ought to be multiplying by the multiplicative identity as we do for other problems of this sort, but instead we are multiplying by [itex]\frac{\frac{1}{x}}{\frac{1}{-\sqrt{x^2}}} = \frac{\frac{1}{x}}{\frac{1}{-x}} = \frac{1}{-1}=-1[/itex].

Not to try to open a whole other can of worms, but it is generally accepted that the square root function, unless otherwise stated, yields its positive root. Just not grasping how we can slap a minus sign there. Sorry if I'm being obtuse; this just seems to be a simple thing that I can't grasp.

Michael Redei
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#4
Dec13-12, 05:42 AM
P: 181

Understanding horizontal asymptotes of non-even rational functions


If ##x## is negative, then both ##\frac1x## and ##\frac1{-\sqrt{x^2}}## are negative, so that the fraction ##
\frac{\frac1x}{-\frac1{\sqrt{x^2}}}
## is equal to 1, and not to -1.
lordofpi
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#5
Dec13-12, 04:56 PM
P: 17
Thanks Michael; I guess that makes sense. The point I couldn't get past, I guess, is that I kept thinking that [itex]\forall x \in \mathbb{R}, \sqrt{x^2} = x \neq -x [/itex]. However, the true definition of the square-root function seems to be that for some number [itex]a \in \mathbb{R}[/itex] we have its square, [itex]a^2 \in \mathbb{R}_{\geq0}[/itex] so that: [tex]\sqrt{a^2} = \begin{cases}
a, & a\ge0\\
-a, & a<0
\end{cases}[/tex]
For example, for [itex]a = -5[/itex], the function [itex]\sqrt{a^2}[/itex]yields [itex]5 = -a[/itex]. In order to obtain the original value of the variable that was squared and then square-rooted when it is taking a negative value, we must multiply it by [itex]-1[/itex]. It all makes perfect sense now. Sorry, I guess I just really needed to do it out. Let me know what you think. Thanks again.
Michael Redei
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#6
Dec13-12, 05:37 PM
P: 181
You've got to the heart of the problem here. I think there's some psychological reason, left over from when we first learn about negative numbers, that makes us think ##-x## must be negative because it has a minus sign. And knowing that square roots are never negative, we have problems with ##\sqrt{x^2} = -x##. How can a positive root be equal to something negative? It seems to make more sense when you re-write that as ##x = -\sqrt{x^2}##, because here you can actually see that ##x## must be something negative (and so ##-x## is positive).

I say "we" because although I've seen 12-year-olds stumble over this problem, so do university students, and also ex-students, like myself.
lordofpi
lordofpi is offline
#7
Dec13-12, 06:29 PM
P: 17
So true. I sort of understand why we are taught a certain way; it is of the greatest benefit for greatest number of students who likely will not pursue anything beyond precalculus ever. However, it becomes a horrible impediment to those of us who will or must go on, who would likely have been better served by a different sort of teaching. I don't have the answers here--just musing. Thanks again for helping me.


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