How do you find the points of continuity for #f(x)=(x-4)/(x^2-16)#?

1 Answer
Gió
May 27, 2015

Apparently this function has two points of discontinuity for #x=+-4# where the denominator becomes zero giving a division by zero that cannot be performed.
One of these two values, however, can be eliminated considering that:
#(x-4)/(x^2-16)=cancel((x-4))/(cancel((x-4))(x+4))=1/(x+4)#
So that only #x=-4# maintains its character of discontinuity while #x=4# is a removable discontinuity.
Graphically:

graph{(x-4)/(x^2-16) [-16.02, 16.02, -8.01, 8.01]}

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