Question Video: Types of Discontinuities Mathematics • Higher Education

Consider the function 𝑓(π‘₯) = π‘₯ cos π‘₯, π‘₯ ≀ πœ‹/2 and 𝑓(π‘₯) = π‘₯ cot π‘₯, π‘₯ > πœ‹/2. What type of discontinuity does the function 𝑓 have at π‘₯ = πœ‹/2, if it has any?

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Video Transcript

Consider the function 𝑓 of π‘₯ is equal to π‘₯ cos of π‘₯ when π‘₯ is less than or equal to πœ‹ over two and 𝑓 of π‘₯ is equal to π‘₯ cot of π‘₯ when π‘₯ is greater than πœ‹ over two. What type of discontinuity does the function 𝑓 have at π‘₯ is equal to πœ‹ over two, if it has any?

The question gives us the piecewise-defined function 𝑓 of π‘₯. And it wants us to decide if the function is continuous or discontinuous at the point where π‘₯ is equal to πœ‹ over two. And if it’s discontinuous, it wants us to find the type of discontinuity.

We recall that we call a function 𝑓 continuous at the point π‘₯ is equal to π‘Ž if it satisfies three conditions. And if any of these conditions fail, depending on which ones and how they fail, we can classify the discontinuity. So we need to check all three of these conditions for our piecewise-defined function 𝑓 of π‘₯ at the point where π‘₯ is equal to πœ‹ over two. So we’ll set π‘Ž equal to πœ‹ over two.

First, we need to check that our function is defined at the point where π‘₯ is equal to πœ‹ over two. From the piecewise definition of our function 𝑓 of π‘₯, we see that when π‘₯ is less than or equal to πœ‹ over two, 𝑓 of π‘₯ is equal to π‘₯ cos of π‘₯. So we substitute π‘₯ is equal to πœ‹ over two into the function π‘₯ cos of π‘₯. This gives us πœ‹ over two multiplied by the cos of πœ‹ over two, which is equal to zero. So our function 𝑓 of π‘₯ is defined at the point where π‘₯ is equal to πœ‹ over two. Another way of saying this is saying that πœ‹ over two is in the domain of our function 𝑓 of π‘₯.

Now we check the second condition. The limit as π‘₯ approaches πœ‹ over two of 𝑓 of π‘₯ must exist. We recall this is equivalent to saying the limit as π‘₯ approaches π‘Ž from the right of 𝑓 of π‘₯ and the limit as π‘₯ approaches π‘Ž from the left of 𝑓 of π‘₯ both exist and are equal. So to check our second condition of continuity, we’ll check the left-hand and right-hand limit of our function 𝑓 of π‘₯ as π‘₯ is approaching πœ‹ over two.

We’ll start with the right-hand limit. Since π‘₯ is approaching πœ‹ over two from the right, we must have that π‘₯ is greater than πœ‹ over two. And we see from our piecewise definition, if π‘₯ is greater than πœ‹ over two, then 𝑓 of π‘₯ is exactly equal to the function π‘₯ cot of π‘₯. This means their limits will be equal in this case. This gives us the limit as π‘₯ approaches πœ‹ over two from the right of π‘₯ cot of π‘₯.

We see we’re trying to evaluate the limit of π‘₯ multiplied by a standard trigonometric function. So we can try to evaluate this limit by using direct substitution. We substitute π‘₯ is equal to πœ‹ over two into the function in our limit. This gives us that our limit is equal to πœ‹ over two multiplied by the cot of πœ‹ over two.

We can directly evaluate the cot of πœ‹ over two to be equal to zero. Or we can rewrite it as the cos of πœ‹ over two divided by the sin of πœ‹ over two, which is equal to zero divided by one, which also gives us zero. So we’ve shown the limit as π‘₯ approaches πœ‹ over two from the right of our function 𝑓 of π‘₯ is equal to zero.

Let’s now evaluate the limit as π‘₯ approaches πœ‹ over two from the left of 𝑓 of π‘₯. Since π‘₯ is approaching πœ‹ over two from the left, we must have that π‘₯ is less than πœ‹ over two. And we see from our piecewise definition of the function 𝑓 of π‘₯, if π‘₯ is less than or equal to πœ‹ over two, our function 𝑓 of π‘₯ is exactly equal to π‘₯ cos of π‘₯.

So if the functions are equal when π‘₯ is less than πœ‹ over two, their limits as π‘₯ approaches πœ‹ over two from the left will also be equal. And we can evaluate the limit as π‘₯ approaches πœ‹ over two from the left of π‘₯ cos of π‘₯ by using direct substitution. Substituting π‘₯ is equal to πœ‹ over two into π‘₯ cos of π‘₯ gives us πœ‹ over two times the cos of πœ‹ over two, which we can evaluate to give us zero.

So we’ve shown the limit as π‘₯ approaches πœ‹ over two from the right exists and the limit as π‘₯ approaches πœ‹ over two from the left exists. And both of these limits are equal. So we’ve shown the second condition for continuity is also true. In fact, the limit as π‘₯ approaches πœ‹ over two of 𝑓 of π‘₯ is equal to zero.

Now we need to check our third condition. The limit as π‘₯ approaches πœ‹ over two of 𝑓 of π‘₯ should be equal to 𝑓 evaluated at πœ‹ over two. In fact, we’ve already evaluated both of these. 𝑓 evaluated at πœ‹ over two is equal to zero. And our limit as π‘₯ approaches πœ‹ over two of 𝑓 of π‘₯ must be equal to the limit as π‘₯ approaches πœ‹ over two from the right and the limit as π‘₯ approaches πœ‹ over two from the left of 𝑓 of π‘₯. These were both equal to zero. This means that our third condition for continuity is also true.

Since all three conditions for continuity are true, we’ve shown that our piecewise function 𝑓 of π‘₯ is continuous at the point π‘₯ is equal to πœ‹ over two. Therefore, we’ve shown that the function 𝑓 does not have a discontinuity at the point where π‘₯ is equal to πœ‹ over two.

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