Question

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

Answer 1

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: