Finding the vertical asymptote of a given function Given any function, how can I find its vertical asymptote? I know that for rationals I can do this by letting the denominator equal to 0. But how about a function like:
$\ln(1-\ln(x))$? You can find the horizontal asymptote of any function by finding the limit of the function as it approaches positive and negative infinity. Is there a generalized algorithm like that for finding the vertical asymptote as well, even when the function is not a rational?
 A: It depends on the type of function.  Here you are assumed to know or figure out or look up that $y=\ln(x)$ has a vertical asymptote at $x=0$, with $\lim\limits_{x\to0+}\ln(x)=-\infty$.  Thus if something else is inside $\ln$, $y=\ln(f(x))$, then there will be a vertical asymptote when what is inside goes to $0$ from the right: where $f(x)\to 0+$.  
So in this case, where you have $y=\ln(f(x))=\ln(1-\ln(x))$, you want to find $a$ such that $1-\ln(x)\to 0+$ as $x\to a$ (possibly only from the right or left).  Can you find such $a$?

$\ln$ also has the property that $\lim\limits_{x\to+\infty}\ln(x)=+\infty$, so $\ln(f(x))$ will go to $+\infty$ where $f(x)\to+\infty$.  For $f(x)=1-\ln(x)$, this occurs where it has its vertical asymptote, because as $x\to0+$, $1-\ln(x)\to +\infty$.

An answer and a comment point out that it is a good idea to go about finding the domain, and check the boundary of the domain.  That is true, but will not in general yield vertical asymptotes; it depends on the particular properties of the functions involved.  It does work for this problem.
A: You simply pick off values of $x$ that are not defined in an orderly manner. For example, in the case of $\ln(1-\ln(x))$, look to the inner nest. By a property of logarithms, $\ln(x)$ is undefined for $x \leq 0$. There is a vertical asymptote at $x=0$, and the function is not defined for $x \leq 0$. Take the next nest, 
$$1-\ln(x) > 0 \Rightarrow \ln(x) <1 \Rightarrow e^{\ln(x)} < e^1 \Rightarrow x < e.$$
So now $x=e$ is another vertical asymptote, and $x<e$.
That is about it. Just collect exclusions for $x$ in an orderly manner. When an exclusion is a boundary as is the case here, it is generally going to be an asymptote.  
A: Can't you just rotate the axis (for a general alogorithm). In any equation, if you switch values of x and y , you obtain sort-of rotated axis. What I mean is, switch x and y in the above equation, then obtain y in terms of x in the new equation, and then use the same techniques you would use for an algorithms for horizontal axis? Let me know if this is clear or not. (Assuming the newly obtained equation can be defined in given interval)
A: I would consider the derivative of your function. Note that while the derivative does not exist at the asymptote point, on the sides of the asymptote point the value of the derivative goes to $\pm \infty $. Which means look for the points where the denominator of the derivative is $0$.
In your case the derivative is: $$\cfrac{-1}{1-\ln(x)}\cdot \cfrac{1}{x}.$$ Can you see where the derivative goes to $\pm \infty$?
As @JonasMeyer pointed out there are other critical points where derivative may go to $\pm \infty$. To avoid those check that the function is undefined at the point where the derivative goes to $\pm \infty$, since, both at vertical tangents and cusp points the function is defined. http://math.uchicago.edu/~vipul/teaching-0910/152/infinitycuspsasymptotes.pdf
