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1.13 Removing Discontinuities

Syllabus
2020
Topic
1.13
Level
—

Learning objectives

—LIM-2.C—Determine values of x or solve for parameters that make discontinuous functions continuous, if possible• LIM-2.C Determine values of x or solve for parameters that make discontinuous functions continuous, if possible.• LIM-2.C.1 If the limit of a function exists at a discontinuity in its graph, then it is possible to remove the discontinuity by defining or redefining the value of the function at that point, so it equals the value of the limit of the function as x approaches that point.• LIM-2.C.2 In order for a piecewise-defined function to be continuous at a boundary to the partition of its domain, the value of the expression defining the function on one side of the boundary must equal the value of the expression defining the other side of the boundary, as well as the value of the function at the boundary.• Enduring understanding LIM-2: Reasoning with definitions, theorems, and properties can be used to justify claims about continuity.

Making a Discontinuity Removable

A discontinuity at x=ax=a can be removed by changing only f(a)f(a) when the finite two-sided limit L=lim⁡x→af(x)L=\lim_{x\to a}f(x) already exists. Define or redefine f(a)=Lf(a)=L; then the limit and the function value agree.

\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)

Suppose g(x)=x2−4x−2g(x)=\dfrac{x^2-4}{x-2} for x≠2x\ne2 and g(2)=kg(2)=k. For x≠2x\ne2, g(x)=x+2g(x)=x+2, so lim⁡x→2g(x)=4\lim_{x\to2}g(x)=4. Choosing k=4k=4 fills the hole and makes gg continuous at 22.

For h(x)=mx+1h(x)=mx+1 when x<2x<2 and h(x)=7h(x)=7 when x≥2x\ge2, continuity at the boundary requires 2m+1=72m+1=7. Thus m=3m=3, making the left-hand limit equal the right-hand limit and h(2)h(2).

Changing one function value cannot remove a jump or vertical-asymptote discontinuity: unequal one-sided limits or an infinite limit mean no finite common limit exists.