Removable & Nonremovable Discontinuity

[nycmathdad]

Hello. I am not too clear concerning the difference between removable and non-removable discontinuity. Look at the picture below. What makes the top function removable and the bottom function non-removable?

 

[Harry_the_cat]

If you factorise the first one, you will see that the factor(x+3) will cancel (but only if x\(\displaystyle \neq\)-3). This does not "remove" the discontinuity. The graph of the first function is still discontinuous at x=-3.
If you factorise the bottom one, no factors will cancel.

 

[nycmathdad]

If you factorise the first one, you will see that the factor(x+3) will cancel (but only if x\(\displaystyle \neq\)-3). This does not "remove" the discontinuity. The graph of the first function is still discontinuous at x=-3.
If you factorise the bottom one, no factors will cancel.

Do you have an example where the rational function has a removable discontinuity?

 

[JeffM]

The first example IS an example of a rational function with a removable discontinuity.

 

[HallsofIvy]

You don't seem to understand what JeffM said. \(\displaystyle f(x)= \frac{x^2- 9}{x- 3}= \frac{(x- 3)(x+ 3)}{x- 3}\) and, as long as x is not 3, we can cancel the "x- 3" in numerator and denominator to get f(x)= x+ 3. We can not do that when x= 3 because then we have 0/0 which has no value. That is, f(x)= x+ 3 for \(\displaystyle x\ne 3\) and f(x) is undefined for x= 3.

That is a discontinuity because because f is not defined at x= 3. it is a removable discontinuity because we can "remove" it by redefining f(3) to be 3+ 3= 6. This new function is identical to x+ 3.

An example of a non-removable function would be f(x)= x+ 3 for ` x \le 0 ` and f(x)= x+ 4 for x> 0. There is no new value for f(0) that will make that continuous.

 

[nycmathdad]

You don't seem to understand what JeffM said. \(\displaystyle f(x)= \frac{x^2- 9}{x- 3}= \frac{(x- 3)(x+ 3)}{x- 3}\) and, as long as x is not 3, we can cancel the "x- 3" in numerator and denominator to get f(x)= x+ 3. We can not do that when x= 3 because then we have 0/0 which has no value. That is, f(x)= x+ 3 for \(\displaystyle x\ne 3\) and f(x) is undefined for x= 3.

That is a discontinuity because because f is not defined at x= 3. it is a removable discontinuity because we can "remove" it by redefining f(3) to be 3+ 3= 6. This new function is identical to x+ 3.

An example of a non-removable function would be f(x)= x+ 3 for $x\le 0$ and f(x)= x+ 4 for x> 0. There is no new value for f(0) that will make that continuous.

You said:

"An example of a non-removable function would be f(x)= x+ 3 for $x\le 0$ and f(x)= x+ 4 for x> 0. There is no new value for f(0) that will make that continuous."

Check you LaTex in that quotation. So, removable discontinuity fills in the hole in the graph of the function by redefining the original function given.

 

[JeffM]

You said:

"An example of a non-removable function would be f(x)= x+ 3 for $x\le 0$ and f(x)= x+ 4 for x> 0. There is no new value for f(0) that will make that continuous."

Check you LaTex in that quotation. So, removable discontinuity fills in the hole in the graph of the function by redefining the original function given.

Yes. One way to think about it is this. If both the right and left limits of f(x) exist and are equal as x approaches a, but f(a) does not exist, redefining f(x) so that f(a) equals that limit removes the discontinuity.

Example

[MATH]f(x) = \dfrac{sin(x)}{x}[/MATH] is continuous everywhere except x = 0.

However, [MATH]\lim_{x \rightarrow 0} \dfrac{sin(x)}{x} = 1.[/MATH]
So if we redefine f(x) as

[MATH]f(x) = \dfrac{sin(x)}{x} \text { if } x \ne 0, \text { and } f(0) = 1,[/MATH]
the redefined function is continuous at x = 0.

 

[nycmathdad]

Yes. One way to think about it is this. If both the right and left limits of f(x) exist and are equal as x approaches a, but f(a) does not exist, redefining f(x) so that f(a) equals that limit removes the discontinuity.

Example

[MATH]f(x) = \dfrac{sin(x)}{x}[/MATH] is continuous everywhere except x = 0.

However, [MATH]\lim_{x \rightarrow 0} \dfrac{sin(x)}{x} = 1.[/MATH]
So if we redefine f(x) as

[MATH]f(x) = \dfrac{sin(x)}{x} \text { if } x \ne 0, \text { and } f(0) = 1,[/MATH]
the redefined function is continuous at x = 0.

Continuity and limits of trig functions is the next section in my textbook. In what way did you redefine sin (x)/x?

 

[JeffM]

Continuity and limits of trig functions is the next section in my textbook. In what way did you redefine sin (x)/x?

I did not redefine sin(x)/x. I redefined f(x).

If we define

[MATH]f(x) = \dfrac{sin(x)}{x}[/MATH], it is undefined if x = 0 because division by 0 is not allowed.

With me to there?

Consequently, f(x) is not continuous at x = 0 by definition because it does not exist at x = 0.

However,

[MATH]\lim_{x \rightarrow 0} f(x) = 1.[/MATH]
This is possible because the definition of limit as x approaches 0 specifies that x does not equal 0.

Thus, if we redefine f(x) to be

[MATH]x \ne 0 \implies f(x) = \dfrac{sin(x)}{x} \text { and } f(0) = 1[/MATH],

we get f(0) = 1 and so f(0) exists and

[MATH]\lim_{x \rightarrow 0} \dfrac{sin(x)}{x} = \lim_{x \rightarrow 0} f(x) = 1 = f(0).[/MATH]
By definition then, the REDEFINED f(x) is continuous at x = 0.

We did nothing to sin(x)/x. We did something to f(x) that eliminated the discontinuity. When we can remove a discontinuity simply by defining a function in a piece-wise manner, we say that the discontinuity is removable.

 

[nycmathdad]

I did not redefine sin(x)/x. I redefined f(x).

If we define

[MATH]f(x) = \dfrac{sin(x)}{x}[/MATH], it is undefined if x = 0 because division by 0 is not allowed.

With me to there?

Consequently, f(x) is not continuous at x = 0 by definition because it does not exist at x = 0.

However,

[MATH]\lim_{x \rightarrow 0} f(x) = 1.[/MATH]
This is possible because the definition of limit as x approaches 0 specifies that x does not equal 0.

Thus, if we redefine f(x) to be

[MATH]x \ne 0 \implies f(x) = \dfrac{sin(x)}{x} \text { and } f(0) = 1[/MATH],

we get f(0) = 1 and so f(0) exists and

[MATH]\lim_{x \rightarrow 0} \dfrac{sin(x)}{x} = \lim_{x \rightarrow 0} f(x) = 1 = f(0).[/MATH]
By definition then, the REDEFINED f(x) is continuous at x = 0.

We did nothing to sin(x)/x. We did something to f(x) that eliminated the discontinuity. When we can remove a discontinuity simply by defining a function in a piece-wise manner, we say that the discontinuity is removable.

This is what I'm talking about. If you (or anyone else) takes time to break down the solution, it becomes clear, in most cases, not all. Section 1.4 is mainly about limits and continuity of trigonometric functions. I am taking my sweet time through Calculus 1 as I will for 2 and 3. Tell me, why do most teachers and students agree that Calculus 2 is harder than 1 and 3?