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Mathematics · Ch 17 — Continuity

Jump Discontinuity

17.1.7

Jump Discontinuity

As seen already in Fig. 8.2, it can happen that both the left-hand limit and the right-hand limit of a function at x=ax=a exist as finite numbers, but the two numbers are different from each other — so the graph visibly "jumps" as xx crosses aa. This situation is called a jump discontinuity:

A function f(x) has a Jump Discontinuity at x=a if lim⁡x→a−f(x) and lim⁡x→a+f(x) both exist but lim⁡x→a−f(x)≠lim⁡x→a+f(x).\textbf{A function } f(x) \textbf{ has a Jump Discontinuity at } x=a \textbf{ if } \lim_{x\to a^-} f(x) \textbf{ and } \lim_{x\to a^+} f(x) \textbf{ both exist but } \lim_{x\to a^-} f(x)\ne\lim_{x\to a^+} f(x).

Illustration 5. Consider f(x)=x2−x−5f(x)=x^2-x-5 for −4≤x<−2-4\le x<-2, and f(x)=x3−4x−3f(x)=x^3-4x-3 for −2≤x≤1-2\le x\le1. At x=−2x=-2, f(−2)=(−2)3−4(−2)−3=−8+8−3=−3f(-2)=(-2)^3-4(-2)-3=-8+8-3=-3 (using the second formula, since −2-2 belongs to that branch). Now

lim⁡x→−2−f(x)=lim⁡x→−2(x2−x−5)=4+2−5=1,lim⁡x→−2+f(x)=lim⁡x→−2(x3−4x−3)=−8+8−3=−3.\lim_{x\to-2^-} f(x)=\lim_{x\to-2}(x^2-x-5)=4+2-5=1,\qquad \lim_{x\to-2^+} f(x)=\lim_{x\to-2}(x^3-4x-3)=-8+8-3=-3. …