To find the left hand limit lim_{x to (pi/2)^-} f(x) we note that for x<pi/2
f(x) = (1-sin^3x)/(3cos^2x)
= ((1-sin x)(1+sin x+sin^2x))/(3(1-sin x)(1+sin x))
= (1+sin x+sin^2x)/(3(1+sin x))
where the last step is legitimate as one approaches pi/2 because (1-sin x) is non-zero as long as x ne pi/2
Thus
lim_{x to (pi/2)^-} f(x) = lim_{x to pi/2} (1+sin x+sin^2x)/(3(1+sin x))=1/2
Hence for continuity, we have
color(red)(a = 1/2)
To figure out the right hand limit we use
(b(1-sinx))/(pi-2x)^2
= b/16(sin^2(x/2)+cos^2(x/2)-2 sin(x/2)cos(x/2))/(x/2-pi/4)^2
= b/16 (sin(x/2)-cos(x/2))^2/(x/2-pi/4)^2
= b/8 (sin(x/2)1/sqrt2-cos(x/2)1/sqrt2)^2/(x/2-pi/4)^2
= b/8 (sin(x/2)cos(pi/4)-cos(x/2)sin(pi/4))^2/(x/2-pi/4)^2
= b/8 sin^2(x/2-pi/4)/(x/2-pi/4)^2
Hence the right hand limit is
lim_{x to (pi/2)^+} f(x) = lim_{x to (pi/2)} b/8 sin^2(x/2-pi/4)/(x/2-pi/4)^2
qquad = b/8 lim_{(x/2-pi/4) to 0} (sin(x/2-pi/4)/(x/2-pi/4))^2 = b/8
Thus for f(x) to be continuous, we must also have
b/8 = 1/2 implies color(red)( b=4)
