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)(adsbygoogle = window.adsbygoogle || []).push({});

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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# Understanding horizontal asymptotes of non-even rational functions

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