Finding asymptotes to general functions

Question

I stumbled upon this problem and didn't quite understand why the solution worked. I have two questions:

1. Why is it that the taylor series is enough to find the oblique asymptote? Specifically, why do the first two terms suffice? It seems like that may not work for all functions.
2. From pre-calculus I recall finding asymptotes of rational functions, but how do I find vertical, horizontal, and oblique asymptotes of a more general function (such as an exponential or square root, etc.)?

Answer 1

In general, the Taylor Series is the last thing you should be looking at.

In fact, the Taylor Series was not looked at; they calculated the Laurent Series. The point is that:

$$\operatorname{e}^y = 1 + y + \frac{1}{2}y^2 + \frac{1}{3!}y^3 + \cdots $$

Replacing $y$ with $2/x$ gives:

\begin{array}{ccc} \operatorname{e}^{2/x} &=& 1 + \frac{2}{x} + \frac{1}{2}\left(\frac{2}{x}\right)^2 + \frac{1}{3!}\left(\frac{2}{x}\right)^3 + \cdots \\ \\ &=& 1 + \frac{2}{x} + \frac{2}{x^2} + \frac{4}{3x^3} + \cdots \end{array}

The actual function under consideration was $x\operatorname{e}^{2/x}+1$ and so we have:

\begin{array}{ccc} x\operatorname{e}^{2/x}+1 &=& x\left(1 + \frac{2}{x} + \frac{2}{x^2} + \frac{4}{3x^3} + \cdots\right) + 1 \\ \\ &=& \left(x + 2 + \frac{2}{x} + \frac{4}{3x^2} + \cdots\right) + 1 \\ \\ &=& x + 3 + \frac{2}{x} + \frac{4}{3x^2} + \cdots \end{array}

The point here is that as $x$ gets larger, anything with an $x$ in the denominator gets smaller: $1/x \to 0$ as $x \to \infty$. So as $x \to \infty$ we have $2/x \to 0$, $4/3x^2 \to 0$, etc. All of "the tail" gets smaller and smaller very quickly. The only terms that we are left with are the $x$ and the $3$.

There is a general definition for an asymptote. They don't even need to be lines.

Answer 2

1) In general it holds for every a that $$\lim\limits_{n->{\infty}} \frac{a}{n} = 0$$. Therefore, all the terms except for the first two disappear.

2) It works in exactly the same way. Just take the limit to infinity and see what result you obtain. However, to take this limit, you sometimes need tricks such as the Taylor polynomials in your example.

Answer 3

You got to be careful with those replacements in taylor series. You can only do that as long as your computations after the replacement stay within the interval of convergence. For the e-power, that's no big deal, since it converges kind of no matter what, but for arctanx for example, there are limitations. Nevertheless, using series is a clever way to find out about possible asymptotes