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

1 Answer
Dec 21, 2016

Answer:

Vertical asymptotes occur at #x=0, ((2n+1)pi)/2# where #n in NN#

Explanation:

We have

#f(x)=1/x-tanx#

which we can write as;

#f(x)=1/x-sinx/cosx#

There will be vertical asymptotes when any part of a denominator is zero, so in this case:

For #1/x# when #x=0#
For #sinx/cosx# when #cosx=0=> x=pi/2,(3pi)/2,(5pi)/2...#

ie at #x=0, ((2n+1)pi)/2# where #n in NN#

We can see these by looking at the graph:
enter image source here