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


*

*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.

*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.)?


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.
