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OceanSpring
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Is a function with a removable discontinuity considered continuous? I've looked through about 6 calculus texts and none of them really go into any detail.
OceanSpring said:Is a function with a removable discontinuity considered continuous? I've looked through about 6 calculus texts and none of them really go into any detail.
That's one flavor of removable discontinuity. But it's also possible for a function to have a removable discontinuity even if it is defined everywhere. For example,TitoSmooth said:What it means by removal discontinuity is that we can define a point( that does not exist in the function. Thus removing the discontinuity.
MrAnchovy said:There seems to be some confusion between a removable discontinuity (which is within the function's domain) and a removable singularity (which is not). A function with a removable discontinuity is not continuous, but a function with a removable singularity may be: a function is continuous if and only if it is continuous everywhere in its domain, i.e. everywhere it is defined.
To clarify the examples in this thread:
##f(x) = \frac{\sin x}{x}## is defined and continuous everywhere other than 0, hence it is continuous. It has a removable singularity at 0.
##f(x) = \frac{x^2-4}{x-2}## is defined and continuous everywhere other than 2, hence it is continuous. It has a removable singularity at 2.
##g(x) = x+2## is defined and continuous everywhere, hence it is continuous.
$$f(x) = \begin{cases}
x + 2 & \text{if }x \neq 2 \\
0 & \text{if } x = 2 \\
\end{cases}$$
is defined everywhere but has a discontinuity at ##x=2##, hence it is not continuous. If we remove ##x=2## from the domain of ##f(x)## then it becomes continuous in the region of ##x=2## and so the discontinuity is removeable.
It isn't "considered continuous"! That's why they use the word discontinuity.OceanSpring said:Is a function with a removable discontinuity considered continuous? I've looked through about 6 calculus texts and none of them really go into any detail.
OceanSpring said:An example would be the rational function x^2-4/x-2. In its original form it would have a hole at 2. Once its been factored and simplified there is no longer a hole.
WWGD said:For yet another perspective, this function is not defined at 2, if it is a Real-valued function.
WWGD said:Until you assign a value at x=2 , you cannot tell whether it is continuous at x=2 or not; 0/0 is undefined.
WWGD said:So yours is a function from ## \mathbb R-{2} \rightarrow \mathbb R ##, and it is continuous in its domain of definition.
WWGD said:To be pedantic (people often tell me I am ;) ), since it is uniformly continuous in a dense subset of the Reals, it can be extended into a continuous function on the Reals.
MrAnchovy said:Hence 2 is not within the domain of the function, by definition.
If you assign a value at x=2 you create a new function. The statement "continuous at x=2" has no meaning for the original function.
Hence it is continuous - that is the definition of a continuous function. To be absolutely clear, a continuous function is not required to be continuous at a value which is not in its domain(!)
This is true, but it does not change the fact that it is already a continuous function on ## \mathbb R-{2} ##
A removable discontinuity, also known as a removable singularity, occurs when there is a missing point or gap in the graph of a function. This means that the function is undefined at a certain value of the independent variable, but can still be made continuous by filling in the gap with a single point.
A removable discontinuity can be identified by looking for a gap or hole in the graph of the function. This gap will typically be a single point, and the rest of the graph will be continuous on either side of the point. Additionally, the function will be defined everywhere except for at the point of the removable discontinuity.
A removable discontinuity is caused by a missing point in the function's graph. This can happen when there is a factor in the function that cancels out, resulting in a hole in the graph at the corresponding value of the independent variable. It can also occur when the function has a vertical asymptote at that point.
A removable discontinuity can be removed by filling in the gap in the graph with a single point. This can be done by finding the limit of the function as it approaches the point of the removable discontinuity, and then evaluating the function at that point. This will result in a continuous function with no gaps in its graph.
Yes, a function can have multiple removable discontinuities. This can occur when there are multiple factors in the function that cancel out, resulting in multiple gaps in the graph. However, as long as each gap can be filled in with a single point, the function can still be made continuous.