
Zach M. answered 05/07/24
Tutor
New to Wyzant
10+ years of teaching Precalculus in classroom setting
- The removable discontinuities would be anywhere that the limit exists and the function is still discontinuous. Since (cot x)^2 is in the denominator and is 0 at x = pi/2 + pi*n , those would be the values where the function has a removable discontinuity.
- The other way to observe it would be to consider trig identities and how the (cot x)^2 identity will simplify the entire function to: j(x) = (sin x)^4 . The (cos x)^2 on the top and bottom reducing similarly removes the discontinuities made by eliminating the possibility of (cos x)^2 = 0 from the denominator which was going to make j(x) undefined at x = pi/2 + pi*n
- this is based on remembering that (cot x)^2 = (cos x)^2 / (sin x)^2 and thus simplifying the function drastically (which is exactly what happens when a limit exists regardless if the function is discontinuous)