What are the asymptote(s) and hole(s), if any, of # f(x) =sin(pix)/x#?

1 Answer
Dec 16, 2017

Answer:

Hole at #x=0 # and a horizontal asymptote with #y = 0#

Explanation:

First you have to calculate the zero marks of the denominator which in this case is #x# therefore there's a vertical asymptote or a hole at #x = 0#. We aren't sure whether this is a hole or asymptote so we have to calculate the zero marks of the numerator
#<=> sin(pi x) =0#
#<=>pi x = 0 or pi x = pi#
#<=> x = 0 or x = 1#
As you see we have a common zero mark. This means that it's not an asymptote but a hole (with #x=0#) and because #x=0# was the only zero mark of the denominator that means that they're are no vertical asymptotes.

Now we take the #x#-value with highest exponent of the denominator and of the numerator and divide them by each other.
but because there is only one kind of exponent of #x# , the function #f(x)# doesn't change.
#<=> sin(pi x)/x#
Now, if the exponent is bigger in the numerator than of the denominator that means that there's a diagonal or a curved asymptote. Otherwise, there's a straight line. In this case, it's going to be a straight line. Now you divide the a values of the numerator by the a value of the denominator.
#<=> Sin(pi)/1#
#<=> 0/1#
#<=> 0#
#<=> y = 0# #=# the horizontal asymptote