How do you find the x values at which #f(x)=csc 2x# is not continuous, which of the discontinuities are removable?

Question

Isabella Ackerman · Accepted Answer

It depends...

The answer to this question depends on your definition of continuity. A function #f(x)# is continuous at a point #x=a# in its domain if and only if: #lim_(x->a) f(x)" "# exists and is equal to #f(a)#. If it is continuous at every point in its domain then according to at least one definition of continuity, we would say that #f(x)# is continuous and has no discontinuities. By this definition, #f(x) = csc 2x# is continuous everywhere. The domain of #csc 2x = 1/(sin 2x)# is the set of values for which #sin 2x != 0#. So #2x != kpi# and hence #x != (kpi)/2# for any integer #k#. Some authors would say that #f(x) = csc 2x# is discontinuous at the points #x=(kpi)/2# on the grounds that: #lim_(x->(kpi)^+) csc 2x = +oo != -oo = lim_(x->(kpi)^-) csc 2x# and: #lim_(x->(((2k+1)pi)/2)^+) csc 2x = -oo != +oo = lim_(x->(((2k+1)pi)/2)^-) csc 2x# That is, the left and right limits disagree at the points #x=(kpi)/2#. Hence if we consider these points of discontinuity then they are not removable. Note however that these points are not part of the domain. graph{csc(2x) [-10, 10, -5, 5]}

Luke Alvarez · Answer

undefined The function f(x) = csc(2x) is not continuous at the values of x where the sine function, sin(2x), equals zero. These values occur when 2x is an integer multiple of π, so x = nπ/2 for n being an integer.

Among these discontinuities, the removable ones are the values of x where the cosecant function, csc(2x), is undefined. This happens when sin(2x) equals zero, resulting in x = nπ/2 for n being an integer.

To summarize, the x values at which f(x) = csc(2x) is not continuous are x = nπ/2 for n being an integer. Among these, the removable discontinuities occur when x = nπ/2 for n being an integer.