Is it possible that the graph of function has infinitely many vertical asymptotes?
I suppose, that it is not possible, because such function would not exist. But I need to prove it in a math-fashioned-way, and I'm clueless how to do it.
Help would be appreciated. :) Thanks.
Generally, a function has a vertical asymptote at $x$ when it can be expressed as: $f(x)=\frac{a}{g(x)}\mid a\neq g(x)$ and $g(x)=0$, the simplest example of which is $\frac{1}{x}$.
For a function to have infinitely many vertical asymptotes there must be infinitely many values of $x$ for which $g(x)=0$. There are two ways this can happen:
(1.) $g(x)$ is periodic with infinitely many zeros - i.e. $$f(x)=u\frac{1}{per(x)}\mid per:=\sin,\cos,\tan,\mod,\ldots,etc.$$ (2.) $f(x)$ is a sum or product of an infinite series.
The latter case is pretty much the same as the former, as most (if not all) such infinite series have a closed form solution equivalent to (1.), for example: $$\sum_{n=0}^\infty(-1)^n\frac{(2n+1)!}{x^{2n+1}}$$ is just the inverse of the Taylor series for the $\sin$ function.
As TonK commented, $\tan{x}$ is a straightforward example. You could also use $\frac{u}{\sin{x}}$, $\frac{u}{\cos{x}}$, $\Gamma(x)$, etc.