The function is defined by g(x)= cases x+1,& x3 cases Which statement about at is correct?

MATHEMATICS · Single correct · AP · Calculus AB · Limits and Continuity · Limits and Continuity

Problem

The function

g

is defined by

g(x)=\begin{cases}x+1,& x<3\\4,& x=3\\x-2,& x>3\end{cases}

Which statement about

g

x=3

is correct? (a) Both one-sided limits exist and are equal, so

g

is continuous at

x=3

. (LHL = RHL = 4) (b) The one-sided limits are unequal, so the limit does not exist and

g

is discontinuous at

x=3

. (LHL = 4, RHL = 1) (c) The one-sided limits are unequal, but because

g(3)=4

the overall limit equals

4

. Hence

g

is continuous at

x=3

. (Limit taken as the function value) (d) Both one-sided limits exist and are equal, but since they are not equal to

g(3)

g

is not continuous at

x=3

. (LHL = RHL = 4,

g(3)=4

assumed unequal)