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Mathematics · Ch 17 — Continuity

Removable Discontinuity

17.1.8

Removable Discontinuity

Some discontinuities are much milder than a jump: the function's limit at the point does exist, but either the function was never given a value there, or it was assigned a value different from that limit. In either case, simply defining or redefining f(a)f(a) to equal the limit patches the function into a continuous one. This is called a removable discontinuity:

A function f(x) has a discontinuity at x=a, and lim⁡x→af(x) exists, but either f(a) is not defined or lim⁡x→af(x)≠f(a).\textbf{A function } f(x) \textbf{ has a discontinuity at } x=a, \textbf{ and } \lim_{x\to a} f(x) \textbf{ exists, but either } f(a) \textbf{ is not defined or } \lim_{x\to a} f(x)\ne f(a).

In such a case, defining or redefining f(a)f(a) as lim⁡x→af(x)\displaystyle\lim_{x\to a} f(x) makes the new function continuous at x=ax=a; this repaired function is called the removable discontinuity, and if the original function was not defined at aa at all, the new, patched definition is called the extension of the original function.

Illustration 6. Consider f(x)=x2+3x−10x3−8f(x)=\dfrac{x^2+3x-10}{x^3-8}, for x≠2x\ne2; here f(2)f(2) is not defined, since the denominator vanishes at x=2x=2. Factorising, x2+3x−10=(x−2)(x+5)x^2+3x-10=(x-2)(x+5) and x3−8=(x−2)(x2+2x+4)x^3-8=(x-2)(x^2+2x+4), so for x≠2x\ne2,

lim⁡x→2f(x)=lim⁡x→2(x−2)(x+5)(x−2)(x2+2x+4)=lim⁡x→2x+5x2+2x+4=2+54+4+4=712.\lim_{x\to2} f(x)=\lim_{x\to2}\frac{(x-2)(x+5)}{(x-2)(x^2+2x+4)}=\lim_{x\to2}\frac{x+5}{x^2+2x+4}=\frac{2+5}{4+4+4}=\frac{7}{12}. …