Question

Where is `f(x) = cot x` discontinuous?

Answer 1

The function will be discontinuous whenever `x= npi`, where `n` is an integer.

Explanation:

We have :

`f(x) =cosx/sinx`

This will be discontinuous whenever the denominator equals `0`, or

`sinx = 0`

Therefore:

`x = 0 or pi`

For a general expression, the function is discontinuous whenever `x = pin`, where `n` is an integer.

Hopefully this helps!

Answer 2

It depends...

Explanation:

Given:

`f(x) = cot(x)`

The domain of `f(x)` is:

`RR "\" { npi : n in ZZ }`

That is `cot(x)` is defined for any real number apart from integer multiples of `pi`.

`cot(x)` is continuous at every point of its domain.

So it is a continuous function.

But what about `x = npi` for any integer `n` ?

Some authors consider these values of `x` points of discontinuity of the function since they are points at which the function is not continuous (indeed it has vertical asymptotes at these points), but they are not in the domain of `cot(x)`.

So some authors consider these to be points of discontinuity and some don't.

Advanced footnote

One way to attempt to harmonise cases where the denominator of a trigonometric function is - so to speak - zero is to consider trigonometric functions to take values in `RR_oo`, the real projective line, instead of `RR`. Note that `RR_oo` only has a single "point at infinity" `oo` instead of the two points `+-oo` that you may be used to. Then the extended definition of `cot(x)` would have `cot(npi) = oo` for any integer `n`.

With this definition, `cot(x)` is not only defined everywhere, but continuous everywhere too.