Questions on curvilinear asymptotes I just saw curvilinear  asymptote which sort of fascinated me. A little bit of thinking raised two questions for which I couldn't get the the answer by googling.
Is there a general method to find a curvilinear  asymptote?
Can a function have more than one curvilinear  asymptote?
PS: I would be grateful if proofs are cited.
 A: (1)There can be several ways of finding curvilinear asymptotes. Some of them can be in terms of polynomials and at other times it can more complicated.
One thing you can do is take the Laurent series of a function. Lets say we have $f(x)=\sqrt{\left(x^4+5x^3+3x+2\right)}$.
You can take the Laurent series (which is the taylor series) by dividing the leading term by replacing taking $f\left(\frac{1}{x}\right)$. Then factoring the leading term and taking the taylor series at x=0 and replace with $\frac{1}{x}$.
$$\sqrt{\left(\frac{1}{x^4}+\frac{5}{x^3}+\frac{3}{x}+2\right)}$$
$$\sqrt{\left(x^4\right)}\sqrt{\left(1+{5}{x}+{3}{x^3}+{2}{x^4}\right)}$$
Now taking the taylor series at $x=0$ and replacing $x$ with $1/x$
$$\lim_{x\to\infty}(x^2)\left(1+\frac{5}{2}x-\frac{25}{8}{x^2}-\frac{3477}{128}x^3+\frac{24835}{256}x^5...\right)$$
$$\lim_{x\to\infty}(x^2)\left(1+\frac{5}{2x}-\frac{25}{8x^2}-\frac{3477}{128x^3}+\frac{24835}{256{x^5}}...\right)$$
$$\lim_{x\to\infty}\left(x^2+\frac{5}{2}x-\frac{25}{8}-\frac{3477}{128x}+\frac{24835}{256{x^2}}...\right)\approx{x^2+\frac{5}{2}x-\frac{25}{8}}$$
You can use this technique on several of these functions. However, you will not always get a polynomial. For example if you take $\sqrt{\left(x^3+5x^2-10x+7\right)}$ you will end up with $x^{3/2}+\frac{5\sqrt{x}}{2}$. This technique is useful for finding asymptotes in the simplest form possible but when functions become compilcated this method can be tedious.
However, if your, function is in the form of a rational function you could leave an asymptote that is not simple.
Taking, for example, $${\left(\frac{x^3+3x^2+5}{x+3}\right)}^{1/3}\left({\frac{x^4+2x^3+5}{x+5}}\right)^{2/5}$$ using a shortcut technique I found you could divide both the polynomials in the bracket.
If you take the brackets and divide inside you get $(x^2)^{1/3}(x^3-3x^2+15x-75)^{2/5}$.
Now you can take this as an asymptote instead taking another laurent series since that would be painful. However all terms on the donimantor must be filled.
Now taking if you take the highest exponent number inside the brackets of ${\left(x^2\right)}^{1/3}$  multiply by x and so the same thing with $\left({x^3-3x^2+15x-75}\right)^{2/5}$ except you multiply with y. Then you replace x and y with the exponents that are lesser than the lowest high exponent number.
$$2(1/3)+3(2/5)<3$$
$$(2/3)+(6/5)<3$$
$$10/15+18/15<3$$
$$33/15<3$$
Thus $(x^2)^{1/3}(x^3-3x^2+15x-75)^{2/5}$ can be an asymptote. However, these asymptotes can only be used at a specific range. And they're are several rules behind this when using sine functions and adding up polynomials....which I'm still not done with plus the proofs for certain rules are too long.
(2) There can be more than one curvalinear asymptotes as long as the original function minus the curvalinear asymptote goes to zero.
