 ### Removable Discontinuities

The first way that a function can fail to be continuous at a point a is that

 lim f(x) = L exists (and is finite) x --> a

but f(a) is not defined or f(a) L. Discontinuities for which the limit of f(x) exists and is finite are called removable discontinuities for reasons explained below.

#### f(a) is not defined

If f(a) is not defined, the graph has a "hole" at (a, f(a)). This hole can be filled by extending the domain of f(x) to include the point x=a and defining

 f(a) = lim f(x). x --> a

This has the effect of removing the discontinuity.

 If lim f(x) = L but f(a) is not defined x --> a

then the discontinuity at x=a can be removed by defining f(a)=L. As an example, consider the function g(x) = (x2 - 1)/(x - 1). Then g(x) = x+1 for all real numbers except x=1. Since g(x) and x+1 agree at all points other than the objective,

 lim g(x) = lim x+1 = 2. x --> 1 x --> 1

We can "remove" the discontinuity by filling the hole. The domain of g(x) may be extended to include x=1 by declaring that g(1) = 2. This makes g(x) continuous at x=1. Since g(x) is continuous at all other points (as evidenced, for example, by the graph), defining g(x) = 2 turns g into a continuous function.

#### The limit and the value of the function are different.

If the limit as x approaches a exists and is finite and f(a) is defined but not equal to this limit, then the graph has a hole with a point misplaced above or below the hole. This discontinuity can be removed by re-defining the function value f(a) to be the value of the limit.

 If lim f(x) = L but f(a) L x --> a

then the discontinuity at x=a can be removed by re-defining f(a)=L.

As and example, the piecewise function in the second equipment check on the page "Defintion of Continuity" was given by

 { Undefined Unless 0 < x < 1 h(x) = 3 If x=.5 1.5 + 1/(x + .25) 0 < x < 1, x .5 We can remove the discontinuity by re-defining the function so as to fill the hole. In this case we re-define h(.5) = 1.5 + 1/(.75) = 17/6.  /Stage4/Lesson/removable.html