# Can asymptotes be curved?

When I was first introduced to the idea of an asymptote, I was taught about horizontal asymptotes (of form $y=a$) and vertical ones ( of form $x=b$).

I was then shown oblique asymptotes-- slanted asymptotes which are not constant (of the form $y=ax+b$).

What happens, though, if we've got a function such as $$f(x)=e^x+\frac{1}{x}?$$

Is $y=e^x$ considered an asymptote in this example?

Another example, just to show you where I'm coming from, is $$g(x)=x^2+\sin(x)$$-- is $y=x^2$ an asymptote in this case?

The reason that I ask is that I don't really see the point in defining oblique asymptotes and not curved ones; surely, if we want to know the behaviour of $y$ as $x \to \infty$, we should include all types of functions as asymptotes.

If asymptotes cannot be curves, then why arbitrarily restrict asymptotes to lines?

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Of course you can. There's, e.g., the notion of polynomial branch, parabolic branch, etc. but then they might not be called asymptotes. – gniourf_gniourf Jun 28 '14 at 21:32
– David H Jun 28 '14 at 21:40
In a way, the notion of asymptotic analysis generalizes the concept of line asymptotes. This site has an asymptotics tag. – Antonio Vargas Jun 28 '14 at 21:45
I would think that a Taylor series expansion is a pretty good example of a general-purpose usage of a curved line for an asymptote. – fluffy Jun 29 '14 at 1:23

The concept of asymptotes is quite common for curved graphs, although somehow the terminology is not much used outside of the context of lines. The way in which the concept is used is that if one is given a function $f(x)$, it is interesting to study other functions $g(x)$ that are "asymptotic to $f(x)$" in various ways. One meaning of this phrase would be that $$(1) \quad \lim_{x \to +\infty} |f(x)-g(x)|=0$$ which is exactly what "asymptotic" means in the ordinary sense when the graph of $f(x)$ is a line. Another somewhat different notion is that $$(2) \quad \lim_{x \to +\infty} \frac{f(x)}{g(x)} = 1$$ which only really makes sense when $f(x)$ and $g(x)$ are nonzero near $+\infty$. There are many other variations on this concept. This discussion falls under the name of "growth types of functions", which are important in computer science and other places; these notes look like a good basic discussion, for example.

And regarding your question of whether $g(x) = x^2 + \sin(x)$ is asymptotic to $y=x^2$, it is asymptotic in sense (2) but not in sense (1).

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"Asymptote," in my view, essentially refers to some kind of limiting behavior of a function. So for instance, we talk about asymptotic series expansions. So from an analytic geometry perspective, we might think of an "asymptote" as a function or relation that describes how another function approaches it arbitrarily closely. For example, $$f(x) = \frac{x-x^2+x^4}{x^2-1}$$ might be thought of as having a parabolic asymptote and two vertical asymptotes, since for "large" $x$, $f(x) \sim x^2$. But I hesitate to say $f(x) = x^2 + \sin x$ has such an asymptote, because the magnitude of $\sin x$ does not diminish as $x$ gets large.

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Though it is still true that $x^2 + \sin x \sim x^2$ as $x \to \infty$ in the usual sense of $\sim$. – Antonio Vargas Jun 28 '14 at 21:48
Yes, of course. – heropup Jun 28 '14 at 21:50
@heropup Is it the fact that $\sin(x)$ oscillates for large $x$ that means that $y=x^2$ wouldn't be considered an asymptote in my case, or is it the fact that $\sin(x)$ doesn't decay to zero? – alexqwx Jun 28 '14 at 21:54
What about $x^2 + \frac{\sin x}{x}$? Would you say that had an asymptote in the same way that the first function you gave had horizontal/vertical asymptotes? – Zubin Mukerjee Jun 28 '14 at 22:18
@ZubinMukerjee In that $\frac{\sin{x}}{x} \rightarrow 0$, yes. – ClickRick Jun 29 '14 at 3:29

terminology is up to you. However, it is useful, when graphing rational functions, to realize that they are essentially polynomial (or the reciprocal of a polynomial) for large absolute values of the argument. Graph $$y = \frac{x^5 - 7}{x^3 - 12 x},$$ for large $|x|,$ $y$ is pretty much $x^2.$

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