We need to check the limit of each function at a=pi/2 and see if it exists and is finite. If the limit does not exist or is infinite, then the function is discontinuous at a=pi/2.
1. f(a) = sin(a)
lim (a->pi/2) sin(a) = sin(pi/2) = 1
The limit exists and is finite, so f(a) is continuous at a=pi/2.
2. g(a) = 1/(a-pi/2)
lim (a->pi/2+) 1/(a-pi/2) = +infinity
lim (a->pi/2-) 1/(a-pi/2) = -infinity
The limits are infinite and have opposite signs, so g(a) is discontinuous at a=pi/2.
3. h(a) = sqrt(cos(a))
lim (a->pi/2+) sqrt(cos(a)) = 0
lim (a->pi/2-) sqrt(cos(a)) = does not exist
The limit from the right is finite, but the limit does not exist from the left. Thus, h(a) is discontinuous at a=pi/2.
which of the following functions are discontinuous at a=pi/2?
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