Removable discontinuity Give an example of a function f (x) th...

Question

Removable discontinuity Give an example of a function f (x) that is continuous for all values of x except x = 2, where it has a removable discontinuity. Explain how you know that f is discontinuous at x = 2, and how you know the discontinuity is removable.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says construct a function H of X that is continuous everywhere except at X equal to 1, where there is a removable discontinuity. Determine the value that H1 should be to remove the discontinuity. We're given 4 possible choices as our answers. For choice A, we have H X is equal to the quantity of X2 minus 1 in quantity, divided by the quantity of X minus 1 in quantity. And H1 is equal to 0. For choice B, we have H of X is equal to the quantity of X2 + 1 in quantity, divided by the quantity of X minus 1 in quantity, and H 1 is equal to 2. For choice C, we have H of X that's equal to the quantity of X minus 1 in quantity, divided by the quantity of X2 minus 1 in quantity, and H1 is equal to 1 divided by 2. And for choice D we have H of X is equal to the quantity of X minus 1 in quantity, divided by the quantity of X2 + 1 in quantity, and H 1 is equal to minus 2. So the first part of this problem was just to construct a function H of X that is continuous everywhere except at X equals 1, where there is a removable discontinuity. So, the easiest way to create a function that is continuous everywhere, except at a single point, is to have a rational expression, where the denominator is equal to 0, when X is equal to 1. So, we need a function, H of X, that is of the form, that is equal to some quantity, divided by the quantity of X minus 1. Since X minus 1 is equal to 0, when X is equal to 1. Now, for this to have a removable discontinuity, in the numerator, we're also going to want to have a factor of X minus 1, which could eventually cancel out later. And the simplest expression that we could have in the numerator, that would be continuous everywhere is a polynomial. So, to create a polynomial, we'll multiply our factor of X minus 1 in the numerator by the quantity of X + 1. So multiplying everything out, we'll have H of X. is equal to the quantity of X2 minus 1 in quantity, divided by the quantity of X minus 1. And so this expression for H of X satisfies all the conditions that were laid out in the problem. So, now we need to do is determine the value that H of one should be in order to remove the discontinuity. So, for this, we're going to look at the limit. As X approaches 1. Of H of X And this is going to be equal to the limit, as X approaches 1, and here we're going to simplify our expression, allowing the factor of X minus 1 in the numerator to cancel out with the factor of X minus 1 in the denominator, and when I'm left with is the quantity of X + 1. And I can evaluate this limit by direct substitution, so this is going to be equal to 1 + 1, which is equal to 2. So that means that H of 1. Should be equal to 2 to remove the discontinuity. So that is the answer to the second half of this problem. So now that we have both halves of our problem answered, we can look at our answer choices and we see that the correct answer choice actually corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

Removable discontinuity Give an example of a function f (x) that is continuous for all values of x except x = 2, where it has a removable discontinuity. Explain how you know that f is discontinuous at x = 2, and how you know the discontinuity is removable.

Key Concepts

Removable Discontinuity

Continuity

Limit

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