A Question About Removable Discontinuity

[dlin320]

Hi guys,
Can anyone help me solve this problem?
By definition, removable discontinuity is when the two-sided limit exists, but isn't equal to the function's value.
In the attached file, the last choice has two limits, the left limit is -4 when x = 2, the right limit is 7 when x = 2. so the two-sided limit doesn't exist. Therefore, this function should not have a removable discontinuity. Am I correct?

Thanks

 

[BigBeachBanana]

Hi guys,
Can anyone help me solve this problem?
By definition, removable discontinuity is when the two-sided limit exists, but isn't equal to the function's value.
In the attached file, the last choice has two limits, the left limit is -4 when x = 2, the right limit is 7 when x = 2. so the two-sided limit doesn't exist. Therefore, this function should not have a removable discontinuity. Am I correct?

Thanks

That's not quite right.

A discontinuity removable at a point [imath]x=a[/imath] if the [imath]\lim_{x \to a} f(x)[/imath] exists and this limit is finite. There are two types of removable discontinuities:
1. The function is undefined at [imath]x=a[/imath]
2. The value of the function at [imath]x=a[/imath] does not match the limit.

 

[JeffM]

Hi guys,
Can anyone help me solve this problem?
By definition, removable discontinuity is when the two-sided limit exists, but isn't equal to the function's value.
In the attached file, the last choice has two limits, the left limit is -4 when x = 2, the right limit is 7 when x = 2. so the two-sided limit doesn't exist. Therefore, this function should not have a removable discontinuity. Am I correct?

Thanks

See the answer of BBB (# 2). But you are correct that a discontinuity is not removable if the limits from both the left and right are finite but not equal. This is entailed, but not explicitly stated, by BBB's comment because no limit exists if the limits from the left and right are not equal. That is, we have one definition dependent on a different definition.

 

[Dr.Peterson]

Hi guys,
Can anyone help me solve this problem?
By definition, removable discontinuity is when the two-sided limit exists, but isn't equal to the function's value.
In the attached file, the last choice has two limits, the left limit is -4 when x = 2, the right limit is 7 when x = 2. so the two-sided limit doesn't exist. Therefore, this function should not have a removable discontinuity. Am I correct?

Thanks

Correct.

So, which of the choices do have a removable discontinuity?

 

[BigBeachBanana]

That's not quite right.

I should clarify this statement. I was referring to your selected answer.

 

[dlin320]

Correct.

So, which of the choices do have a removable discontinuity?

The answer key shows the last choice has a removable discontinuity. But I don't agree with that.

 

[dlin320]

I should clarify this statement. I was referring to your selected answer.

The selected answer is the key.

 

[dlin320]

I think the official answer is incorrect. Somebody made a mistake there Choices a and c describe essentially the same function, but the first has a removable discontinuity at x=2 and the latter has removed that discontinuity.

 

[Steven G]

The answer key is wrong! You are down to three choices, which do you pick and why?

 

[dlin320]

The answer key is wrong! You are down to three choices, which do you pick and why?

I think the official answer is incorrect. Somebody made a mistake there Choices a and c describe essentially the same function, but the first has a removable discontinuity at x=2 and the latter has removed that discontinuity. so it should be a.

 

[Dr.Peterson]

The answer key shows the last choice has a removable discontinuity. But I don't agree with that.

I agree with you in disagreeing with the answer key! But that doesn't answer my question.

As for what to do about the wrong answer, you should contact your teacher (and/or the source, whatever it is) to let them know about it. Don't believe everything in answer keys.

 

[pka]

The answer key shows the last choice has a removable discontinuity. But I don't agree with that.

You may not agree with this but it is a fact: [imath]\large{\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - x - 2}}{{x - 2}} = 3}~.[/imath]
The statement that [imath]\mathop {\lim }\limits_{x \to {x_0}} f(x) = L[/imath] means that if [imath]x\approx x_0\text{ BUT }x\ne x_0[/imath] then [imath]f(x)\approx L[/imath].
In other words, if [imath]x\text{ is close to }x_0\text{ then }f(x)\text{ is close to }L[/imath].
Do you see that [imath]\dfrac{{{x^2} - x - 2}}{{x - 2}} = \dfrac{{(x + 1)(x - 2)}}{{(x - 2)}}=(x+1)\text{ if }x\ne 2~?[/imath]
The function[imath]f(x)=\dfrac{{{x^2} - x - 2}}{{x - 2}}[/imath] is not continuous at [imath]x=2[/imath] because it is not defined there.
At the same time it has a limit of [imath]\bf 3[/imath] there. Hence the third option is continuous at [imath]x=2[/imath].
But one could argue that it not a removeable discontinuity because there is nothing to remove.

 

[dlin320]

You may not agree with this but it is a fact: [imath]\large{\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - x - 2}}{{x - 2}} = 3}~.[/imath]
The statement that [imath]\mathop {\lim }\limits_{x \to {x_0}} f(x) = L[/imath] means that if [imath]x\approx x_0\text{ BUT }x\ne x_0[/imath] then [imath]f(x)\approx L[/imath].
In other words, if [imath]x\text{ is close to }x_0\text{ then }f(x)\text{ is close to }L[/imath].
Do you see that [imath]\dfrac{{{x^2} - x - 2}}{{x - 2}} = \dfrac{{(x + 1)(x - 2)}}{{(x - 2)}}=(x+1)\text{ if }x\ne 2~?[/imath]
The function[imath]f(x)=\dfrac{{{x^2} - x - 2}}{{x - 2}}[/imath] is not continuous at [imath]x=2[/imath] because it is not defined there.
At the same time it has a limit of [imath]\bf 3[/imath] there. Hence the third option is continuous at [imath]x=2[/imath].
But one could argue that it not a removeable discontinuity because there is nothing to remove.

I chose A as the correct answer. C has removed that discontinuity.

 

[dlin320]

I agree with you in disagreeing with the answer key! But that doesn't answer my question.

As for what to do about the wrong answer, you should contact your teacher (and/or the source, whatever it is) to let them know about it. Don't believe everything in answer keys.

Finally, I chose A.