How do I find the vertical asymptotes of #f(x) = tanπx#?

1 Answer
Sep 7, 2015

Answer:

There are infinite (countable) number of asymptotes described by the following expression for #x#:
#x = 1/2 + N#, where #N# - any integer number.

Explanation:

By definition, the vertical asymptote of a function is a vertical line on the coordinate plane that intersects the X-axis at a point where the value of a function is undefined and is infinitely increasing to #+oo# or infinitely decreasing to #-oo# as its argument #x# approaches to this point.

Since the definition of a function #tan(phi)# is #sin(phi)/cos(phi)#, function #tan()# has asymptotes wherever #cos()# is equal to zero.

So, to get an asymptotes of function #tan(pi x)#, we have to resolve an equation
#cos(pi x)=0#

As we know, function #cos(phi)# represents an abscissa (X-coordinate) of a point on a unit circle that is an endpoint of a vector at an angle #phi# with a positive direction of the X-axis. So, it equals to zero when this vector is either vertically directed up or down along the Y-axis, which corresponds to angles #pi/2# and #-pi/2#. Adding periodicity, we can say that an angle must be equal to #pi/2 + pi N#, where #N# - any integer number.

At #N=0# we get #x=pi/2#.
At #N=-1# we get #x=-pi/2#.
At #N=1# we get #x=3pi/2# (same angle as #-pi/2#).
At #N=-2# we get #x=-3pi/2# (same angle as #pi/2#).
etc.

So, we have a solution for our equation:
#pi x = pi/2 + pi N# or
#x = 1/2 + N#,
where #N# - any integer number.

These values of #x# are those points where function #tan(pi x)# has vertical asymptotes.