What kind of discontinuity is this piecewise function? Removable or jump?
f(x) ={ (2x^2 - 5x - 3)/(x-3) if x does not equal 3
............6............................if x = 3
jump.
f(x) = 2x+1 everywhere except x=3.
Since the limit from both sides is 7, but f(3)=6, there's no way to remove that break.
I thought that since the limit exists, but does not equal f(3) it would be removable
It is only removable if f(x) is defined to be the same as the limit value. If f(x) were 7 at x=3, then that would plug the hole. But since it is defined to be a different value, it creates a jump which cannot be joined to fill the hole.
Thank you so much for explaining that!
To determine the type of discontinuity in the piecewise function, we need to evaluate the limit of the function as x approaches the point of interest, which in this case is x = 3.
First, let's calculate the limit of the function as x approaches 3 from the left side, denoted as f(3-):
lim(x→3-) (2x^2 - 5x - 3)/(x - 3)
To find this limit, we can substitute x = 3 into the expression for f(x):
f(3-) = (2(3)^2 - 5(3) - 3)/(3 - 3) = (18 - 15 - 3)/(0) = 0/0
Since we obtained an indeterminate form of 0/0, we cannot evaluate the limit without further simplification.
Next, let's calculate the limit of the function as x approaches 3 from the right side, denoted as f(3+):
lim(x→3+) (2x^2 - 5x - 3)/(x - 3)
Again, we substitute x = 3 into the expression for f(x):
f(3+) = (2(3)^2 - 5(3) - 3)/(3 - 3) = (18 - 15 - 3)/(0) = 0/0
The right-hand limit also yields an indeterminate form of 0/0.
Since both the left-hand and right-hand limits approached 0/0, this indicates that the function has a removable discontinuity at x = 3. In other words, the function has a hole at x = 3 where the original function is not defined, but it can be filled by evaluating the limit and assigning a value.