In order that a real function undefined at x = π/2 can be redefined
with a value f(π/2) and be continuous at x = π/2, it must:
A. be defined and continuous on both sides of π/2. That is, defined both
on (π/2-a,π/2) and (π/2,π/2+b), for some nonnegative a and b, and
B. have the same limits on both sides of π/2. That is
, where both limits exist and are finite.
As it turns out none of the functions listed fit both requirements.
None approach the same finite limit on both sides of π/2.
(1) is not defined left of π/2
(2) and (3) have vertical asymptotes at π/2.
(4) is not defined immediately to the right of π/2,
Now if you mis-typed the right parenthesis on
2) abs(sin(2x))/(2x- π)
and it should have been
2) abs(sin(2x)/(2x- π))
where both the numerator and denominator are included in the
absolute value, then that would have satisfied both requirements.
The limits on both sides of π/2 would be +1 and the function could
be redefined with f(π/2)=1, and the function would be continuous at
π/2.
If that is what you meant, or if you made some other typo, please
re-post.
Edwin
