Find all the values of x for which f(x)= 3/ (1-cos2x) fails to be continuous
There are several theorems about continuous functions that we need to use. First, every polynomial is continuous everywhere, so 2x is a continuous function, as is the constant function y=3 in the numerator. Also, cos u is continuous for all u, and the composition of two continuous functions is continuous, making cos(2x) continuous everywhere. Differences of continuous functions are also continuous, so 1-cos(2x) is continuous everywhere.
You may not need to provide all that detail for your answer, but the key theorem we need is that the quotient of two continuous functions is continuous everywhere except where the denominator is 0, so our points of discontinuity will occur whenever
`1-cos(2x)=0` , or equivalently, `cos(2x)=1`. But the value of cosine is 1 only for angles that are an integer multiple of `2pi`, so our solutions are `2x=2pi*n` , or `x=pi*n` where `n` is an integer. Here is the graph:
The function is discontinuous at the infinitely many points `x=pi*n` , where `n` is an integer.