# Find vertical asymptote?

Normally to find the vertical asymptote of function like $y=(\ln x)/x$, we would just say 'the vertical asymptote occurs when the denominator equals zero'. This works fine for most functions, however how do we determine what the vertical asymptote is for a function like $y=\ln x + 1/x$? There is no common denominator, yet there is still an asymptote. What is an easy way of calculating the asymptote for functions like these?

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You try to find values where your function is undefined. In this case, you know that zero is a "bad" value for both $1/x$ and $\ln\,x$, so... –  Ｊ. Ｍ. Nov 6 '11 at 5:34
You can rewrite the function so it has a common denominator, for one thing: $$y = \ln(x) + \frac{1}{x} = \frac{x\ln x + 1}{x}.$$Note that you can have vertical asymptotes without denominators (like in $\ln x$) and you can have zeros in the denominator with no vertical asymptotes (like in $\frac{x^2-2x+1}{x-1}$). –  Arturo Magidin Nov 6 '11 at 5:38
Thanks Arturo! So how do I know when something will have an asymptote but not a denominator, or when it will have a denominator but not any asymptote? What do I look for? –  Parachuting Panda Nov 6 '11 at 5:50
@ParachutingPanda: Zeros of the denominator are good places to start, but you need to see what happens to the function as you approach that value (the reason my second example does not have an asymptote is that as you approach $1$, the value of the function just approaches $x-1$, it doesn't "blow up"). For more general functions, you should look for potential problems in places where the function is not continuous; just form staring at them it may be difficult, generally you know a bit about the functions that make up the one you are looking at (such as knowing about $\ln x$ here). –  Arturo Magidin Nov 6 '11 at 6:32
@ParachutingPanda Where would $\ln(u)$ have a vertical asymptote? Once you have those values of $u$, find the values of $x$ such that $x^2-1=u$ (for those values). –  process91 Nov 6 '11 at 17:30