Editing Asymptote (section)

===Vertical asymptotes===
The line ''x'' = ''a'' is a ''vertical asymptote'' of the graph of the function {{nowrap|1=''y'' = ''&fnof;''(''x'')}} if at least one of the following statements is true:

# <math>\lim_{x \to a^{-}} f(x)=\pm\infty,</math>
# <math>\lim_{x \to a^{+}} f(x)=\pm\infty,</math>

where <math>\lim_{x\to a^-}</math> is the limit as ''x'' approaches the value ''a'' from the left (from lesser values), and <math>\lim_{x\to a^+}</math> is the limit as ''x'' approaches ''a'' from the right.

For example, if ƒ(''x'') = ''x''/(''x''–1), the numerator approaches 1 and the denominator approaches 0 as ''x'' approaches 1. So
:<math>\lim_{x\to 1^{+}}\frac{x}{x-1}=+\infty</math>
:<math>\lim_{x\to 1^{-}}\frac{x}{x-1}=-\infty</math>
and the curve has a vertical asymptote ''x'' = 1.

The function ''ƒ''(''x'') may or may not be defined at ''a'', and its precise value at the point ''x'' = ''a'' does not affect the asymptote.  For example, for the function

:<math>f(x) = \begin{cases} \frac{1}{x} & \text{if } x > 0, \\ 5 & \text{if  } x \le 0. \end{cases}</math>

has a limit of +∞ as {{nowrap|''x'' &rarr; 0<sup>+</sup>}}, ''ƒ''(''x'') has the vertical asymptote  {{nowrap|1=''x'' = 0}}, even though ''ƒ''(0)&nbsp;=&nbsp;5. The graph of this function does intersect the vertical asymptote once, at (0, 5). It is impossible for the graph of a function to intersect a vertical asymptote (or [[vertical line test|a vertical line in general]]) in more than one point. Moreover, if a function is [[continuous function|continuous]] at each point where it is defined, it is impossible that its graph does intersect any vertical asymptote.

A common example of a vertical asymptote is the case of a rational function at a point x such that the denominator is zero and the numerator is non-zero.

If a function has a vertical asymptote, then it isn't necessarily true that the derivative of the function has a vertical asymptote at the same place. An example is
:<math>f(x) =  \tfrac 1x + \sin(\tfrac 1x)\quad</math> at <math>\quad x=0</math>.

This function has a vertical asymptote at <math>x=0,</math> because 
:<math>\lim_{x\to0^+} f(x) = \lim_{x\to0^+}\left(\tfrac 1x + \sin\left(\tfrac 1x\right)\right) = +\infty,</math>

and

:<math>\lim_{x\to0^-} f(x) = \lim_{x\to0^-}\left(\tfrac 1x + \sin\left(\tfrac 1x\right)\right) = -\infty</math>.

The derivative of <math>f</math> is the function
:<math>f'(x)=\frac{-(\cos(\tfrac 1x) + 1)}{x^2}</math>.

For the sequence of points
:<math>x_n=\frac{(-1)^n}{(2n+1)\pi},\quad</math> for <math>\quad n=0,1,2,\ldots</math>

that approaches  <math>x=0</math> both from the left and from the right, the values <math>f'(x_n)</math> are constantly <math>0</math>. Therefore, both [[one-sided limit]]s of <math>f'</math> at <math>0</math> can be neither <math>+\infty</math> nor <math>-\infty</math>. Hence <math>f'(x)</math> doesn't have a vertical asymptote at <math>x=0</math>.