Question

What's the difference between jump and removable discontinuity?

Answer 1

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]}