Mathematics

Mathematics

Problem

The function m(x)={sin⁡x,x<0cos⁡x,x≥0m(x)= \begin{cases} \sin x, & x<0\\ \cos x, & x\ge 0 \end{cases} exhibits a discontinuity at x=0x=0. Which of the following best explains why this discontinuity is classified as a jump discontinuity? a) Both one-sided limits are equal, so lim⁡x→0m(x)=0\displaystyle\lim_{x\to 0} m(x)=0, but this limit is not equal to m(0)=1m(0)=1. b) The left-hand and right-hand limits at x=0x=0 both exist but are not equal. c) Both one-sided limits diverge to infinity as x→0x\to 0. d) The right-hand limit exists at x=0x=0, but the left-hand limit does not, causing the discontinuity.

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