Function with a removable discontinuity

[OceanSpring]

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.

 

[pwsnafu]

Continuous where? A function can have a removable discontinuity at $x=a$, but have a jump discontinuity at $x=b$, so it wouldn't be a continuous function. On the other hand $\frac{\sin x}{x}$ is defined and continuous everywhere other than 0, hence it is continuous.

 

[OceanSpring]

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.

 

[pwsnafu]

$f(x) = \frac{x^2-4}{x-2}$ and $g(x) = x+2$ are different functions. The former has domain of $\mathbb{R}\setminus\{2\}$ (with a removable singularity at $x=2$) the latter has domain $\mathbb{R}$ (with no singularity). Both functions are continuous.

Lastly, notation: "x^2-4/x-2" means $x^2-\frac{4}{x}-2$; "(x^2-4)/(x-2)" means $\frac{x^2-4}{x-2}$.

 

[TitoSmooth]

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.

A function is continuous.

1. The left hand limit=right habd limit
At a point a.

2. The function f (x)=a is defined at the point a. (In the domain).


What it means by removal discontinuity is that we can define a point( that does not exist in the function. Thus removing the discontinuity.

 

[jbunniii]

What it means by removal discontinuity is that we can define a point( that does not exist in the function. Thus removing the discontinuity.

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,
$$f(x) = \begin{cases}
x + 2 & \text{if }x \neq 2 \\
0 & \text{if } x = 2 \\
\end{cases}$$
is defined for all $\mathbb{R}$ and has a removable discontinuity at $x=2$, hence is discontinuous at that point. If we redefine $f(2) = 4$ then it becomes a continuous function. This is why we say that the discontinuity is removable. But this example shows that the answer to the OP's question is no: the above function is not continuous.

 

[pbuk]

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.

 

[symbolipoint]

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.

Great Try, but even now, I'm still confused. This continuity stuff and the epsilon-delta proofs were the difficult parts of Calculus 1. I wish I could understand this more clearly - maybe I could make better progress in Mathematics. (Although succeeded in Calculus 1, 2, 3; many many years ago with tough long effort).

 

[HallsofIvy]

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.

It isn't "considered continuous"! That's why they use the word discontinuity.

Can you give a reference for a textbook that says "a function with a removable discontinuity is considered continuous"?

What is true is that if f(x)= g(x) for all x except x= a, then \lim_{x\to a} f(x)= \lim_{x\to a} g(x). But that is NOT saying that f(x) and g(x) are the same function.

 

[WWGD]

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.

For yet another perspective, this function is not defined at 2, if it is a Real-valued function. Until you assign a value at x=2 , you cannot tell whether it is continuous at x=2 or not; 0/0 is undefined . Like many others said, the function _can be defined_ at x=2 as to be made continuous there, but, as it is, it is not even defined there using your rule. So yours is a function from $\mathbb R-{2} \rightarrow \mathbb R$, and it is continuous in its domain of definition. 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.

 

[pbuk]

For yet another perspective, this function is not defined at 2, if it is a Real-valued function.

Hence 2 is not within the domain of the function, by definition.

Until you assign a value at x=2 , you cannot tell whether it is continuous at x=2 or not; 0/0 is undefined.

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.

So yours is a function from $\mathbb R-{2} \rightarrow \mathbb R$, and it is continuous in its domain of definition.

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(!)

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.

This is true, but it does not change the fact that it is already a continuous function on $\mathbb R-{2}$

 

[WWGD]

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 $\mathbb R-{2}$

Sorry, I don't get any of your points; you seem to be saying what I said in different words. I never said 2 was in the domain, I never said the function was either defined at x=2 nor that it was continuous there. And I did not say that f was not continuous on $\mathbb R-2$.