Horizontal and Oblique Asymptotes - A more advanced but universal approach

[RuiAce]

By that, I mean an approach that's probably too much for you to expect to know, but isn't subject to limitations.

When you're a 3U approach, you start to sketch curves other than \(y = \frac1x\), that have both horizontal and vertical asymptotes in them. Vertical asymptotes are easy enough; just look out for when you make black holes (i.e. divide by 0). When you ONLY had horizontal asymptotes, they weren't bad either.

A case of a horizontal asymptote

Which is, in fact, CORRECT.

But then they introduced oblique asymptotes. This messed things up

A case of an oblique asymptote, where the problems are

Which is fair enough, because an oblique asymptote DOES imply that the graph eventually explodes. But it doesn't tell us any information about the asymptote itself, i.e. just HOW FAST does it explode.

So instead of dividing by the highest power of \(x\), we can cover up the "negligible" terms. We go over this again soon.


Now this isn't far off, but it is still WRONG. Because the actual asymptote is \( y = \frac{x}{2} + \frac{1}{8} \)

Essentially, very simple tricks that work for horizontal asymptotes won't work for oblique asymptotes. The problem is more or less the fact that a horizontal asymptote essentially means that the graph tends towards a constant. A constant doesn't grow, because it's just a fixed number.

An oblique asymptote, on the other hand, involves growth. You grow ever so close to a linear fashion, and the above approaches only determine the gradient of that straight line. It doesn't tell you anything about whether it's shifted up or down, and hence we introduce the 3U preferred method.

The same case of the oblique asymptote, but with problems resolved

In actuality, provided \(P(x)\) is a polynomial whose degree is less than \(Q(x)\), another polynomial, then \( \lim_{x\to \infty} \frac{P(x)}{Q(x)} \) will always be 0. The 'divide by highest power of \(x\)' technique will verify this.

But right now, our numerator has a higher degree than the denominator. The main way to rectify this issue is through polynomial long division (or equivalent).

And at this point, the fraction hanging on the end now has a numerator, whose degree is less than the denominator. So we can deduce it goes to 0 as \(x\to \infty\).

Thus, giving us the correct oblique asymptote of \( y = \frac{x}{2} + \frac{1}{8} \)

Now Rui. You've given me a nice technique here and you've just basically told us it works for all of these graphs. What are you getting at here?

Well, what I'm getting at now is the possibility that you don't have a fraction.

Disclaimer: As far as I know of, you should never see this kind of curve in 3U. This question was actually sourced from the 4U section of the forum. It lead me to believe that my techniques I used for 3U had started to become insufficient, but the concepts presented below are not beyond a 3U student's capability; hence why I post it here and not in the 4U forum.

Earlier, I mentioned a cover-up method. Now, what "cover-up" is saying in THIS context is that once \(x\) gets really really large, a certain term will dominate. That dominating term, will be whatever grows the "fastest".

What I mean by "grows the fastest" is this. Consider a quadratic: \(x^2 + x\). The terms that build it up are \(x\) and \(x^2\).
But between the two, \(x^2\) is gonna grow faster than \(x\). Because it's the SQUARE of something that's already growing really fast. So here, \(x^2\) serves as the "dominating" term.

For another example, consider this random mess: \( \frac{1}{100}x^{100} + \frac{999}{2} x^{0.5} + 1000x^{50.5} \). At the end of the day, regardless of the coefficients, \( \frac{1}{100}x^{100} \) will be the dominating term, since it has the highest power on \(x\).
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When we cover-up, we're getting rid of all of the recessive (i.e. dominated) terms. In the first case, that would just be the \(x\). And that makes sense, because \(x^2+x\) will grow at almost the same rate as \(x^2\) does.

Note how above, I say almost. In reality, it'll grow faster because \(x\) is not a constant. If we compared \(x^2\) to \(x^2+1\) on the other hand, then they'd be growing at essentially the same rate, since all the constant does is provide an offset.

The aim of this post, is to show how cover-up can be helpful when we have oblique asymptotes.
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First, go back to the earlier example

Our aim is to show that \(C = \frac{1}{8}\) as above. So the question is how?
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Comparing the two boxed expressions, \( C = \frac18 \) as required.

So cover-up is definitely a longer approach, but it's quite nice in that it'll still work. But of course, for the other example, since we don't have a fraction involving polynomials, our only choice is to use this cover up technique.

Doing that nasty 4U problem

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As it turns out, this is quite hard. Some of you may have seen this before when we dealt with square roots. Our trick was to effectively rationalise the numerator. As an example, you may like to try solving this limit: \( \lim_{x\to \infty} (\sqrt{x+1}-\sqrt{x}) \)

Here, we deal with cubes. So instead of using the difference of two squares or something, I will use the sum and difference of two cubes formulae.
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