Infinite amount of vertical asymptotesIs 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.

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Infinite amount of vertical asymptotesIs 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&#039;m clueless how to do it.

erlentzed · Accepted Answer

Generally, a function has a vertical asymptote at x when it can be expressed as:   f  (  x  )  =      a          g      (      x      )        ∣  a  ≠  g  (  x  ) and g(x)=0, the simplest example of which is       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      1          p      e      r      (      x      )        ∣  p  e  r  :=  sin  ,  cos  ,  tan  ,    mod      ,  …  ,  e  t  c  .(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:      ∑          n      =      0        ∞    (  −  1      )    n              (      2      n      +      1      )      !              x              2        n        +        1            is just the inverse of the Taylor series for the sin function.As TonK commented, tanx is a straightforward example. You could also use       u          sin      ⁡              x            ,       u          cos      ⁡              x            ,   Γ  (  x  ), etc.

Infinite amount of vertical asymptotes 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.

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