The function m(x)= cases x, & x<0 x, & x 0 cases exhibits a discontinuity at

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

Problem

The function

m(x)= \begin{cases} \sin x, & x<0\\ \cos x, & x\ge 0 \end{cases}

exhibits a discontinuity at

x=0

. Which of the following best explains why this discontinuity is classified as a jump discontinuity? (a) Both one-sided limits are equal, so

\displaystyle\lim_{x\to 0} m(x)=0

, but this limit is not equal to

m(0)=1

. (b) The left-hand and right-hand limits at

x=0

both exist but are not equal. (c) Both one-sided limits diverge to infinity as

x\to 0

. (d) The right-hand limit exists at

x=0

, but the left-hand limit does not, causing the discontinuity.