For what values of x, if any, does #f(x) = 1/((x-2)sinx) # have vertical asymptotes?

1 Answer
Mar 13, 2016

Answer:

#x=2#, #x=2pik#, and #x=pi+2pik# where #k in Z#

Explanation:

Vertical asymptotes occur whenever the denominator equals 0. To find them, we simply set the denominator to 0 and solve, like thus:
#(x-2)sinx=0#
#x-2=0# and #sinx=0#
#x=2# and #x=2pik, pi+2pik# where #k in Z# (#k# is an integer)

We have infinitely many vertical asymptotes, at #x=2#, #x=2pik#, and #x=pi + 2pik#. We can see this on the graph of the function below:
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