What's the difference between jump and removable discontinuity?

1 Answer
May 7, 2015

First of all, we are talking about a behavior of a function around some particular value of an argument.

Let's assume our function is represented as #y=f(x)# and its behavior around the value of argument #x=a# is what we are analyzing.

Also assume that the following two limits are properly defined, exist and are finite:
1. Limit of the function as argument #x# approaches a value #a# from the left:
#c=lim_(x->a^-)f(x)#
2. Limit of the function as argument #x# approaches a value #a# from the right:
#d=lim_(x->a^+)f(x)#

If the above limits are equal and a function #f(x)# is defined at #x=a# and its value is the same as these limits, we have no discontinuity.
In other words, the condition for not having a discontinuity is
#lim_(x->a^-)f(x) = lim_(x->a^+)f(x) = f(a)#
Example of a function with no discontinuity is
#y=|x|#:
graph{|x| [-10, 10, -5, 5]}

If the above limits are not equal, we have a jump discontinuity.
In other words, the condition for a jump discontinuity is
#lim_(x->a^-)f(x) != lim_(x->a^+)f(x)#
Example of a function with a jump discontinuity at #x=0# is
#y=-1# for negative #x#,
#y=1# for positive #x# and
#y=0# for #x=0#:
graph{x/|x| [-10, 10, -5, 5]}

Finally, If the above limits are equal but a function #f(x)# is either not defined at #x=a# or is defined, but its value is not the same as these limits, we have a removable discontinuity.
In other words, the condition for a removable discontinuity is
#lim_(x->a^-)f(x) = lim_(x->a^+)f(x)# AND
#f(a)# IS UNDEFINED OR #f(a) != lim_(x->a^-)f(x)#
Example of a function with removable discontinuity is
#y=x^2# everywhere except #x=0# and
function is undefined at #x=0# :
graph{x^3/x [-10, 10, -5, 5]}