# Video: Discussing the Continuity of a Piecewise-Defined Functions Involving Trigonometric Ratios at a Point

Discuss the continuity of the function 𝑓 at 𝑥 = 𝜋/2, given 𝑓(𝑥) = 8 + cos 7𝑥, if 𝑥 < 𝜋/2 and 𝑓(𝑥) = 7 + sin 5𝑥, if 𝑥 ≥ 𝜋/2.

05:26

### Video Transcript

Discuss the continuity of the function 𝑓 at 𝑥 is equal to 𝜋 by two, given that the function 𝑓 of 𝑥 is equal to eight plus the cos of seven 𝑥 if 𝑥 is less than 𝜋 by two, and the function 𝑓 of 𝑥 is equal to seven plus the sin of five 𝑥 if 𝑥 is greater than or equal to 𝜋 by two.

The question gives us a piecewise-defined function 𝑓 of 𝑥 and asks us to decide if the function is continuous at the value 𝑥 is equal to two [𝜋 by two]. Let’s start by recalling what it means for a function to be continuous at a point. We say that a function 𝑓 of 𝑥 is continuous at the value 𝑥 is equal to 𝑎 if the following three conditions are true.

First, 𝑓 evaluated at 𝑎 must be defined. This is the same as saying that 𝑎 is in the domain of our function 𝑓. Second, we must have the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 exists. And it’s worth noting this is the same as saying that both the limit as 𝑥 approaches 𝑎 from the left of 𝑓 of 𝑥 and the limit as 𝑥 approaches 𝑎 from the right of 𝑓 of 𝑥 exist and are equal. Finally, we need the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 to be equal to 𝑓 evaluated at 𝑎. We’re only interested in the continuity of 𝑓 when 𝑥 is equal to 𝜋 by two, so we’ll set 𝑎 equal to 𝜋 by two. So that’s right. 𝑎 is 𝜋 by two in our definition for continuity.

The first thing we need to check is 𝑓 is defined at 𝑥 is equal to 𝜋 by two. And we can do this directly from the piecewise definition of 𝑓 of 𝑥. We can see when 𝑥 is equal to 𝜋 by two, our function 𝑓 of 𝑥 is equal to seven plus the sin of five 𝑥. So 𝑓 evaluated at 𝜋 by two is equal to seven plus the sin of five times 𝜋 by two. And we can evaluate this by using the fact that the sine function is periodic about an interval of two 𝜋. This tells us that sin of five 𝜋 by two is equal to the sin of 𝜋 by two which we know is equal to one. So this tells us 𝑓 evaluated at 𝜋 by two is equal to seven plus one, which is eight. So we’ve shown that 𝑓 evaluated at 𝜋 by two is defined. In other words, 𝜋 by two is in the domain of our function 𝑓 of 𝑥.

Let’s move on to our second condition for continuity. We need to show the limit as 𝑥 approaches 𝜋 by two of 𝑓 of 𝑥 exists. And we’ll do this by showing the limit as 𝑥 approaches 𝜋 by two from the left of 𝑓 of 𝑥 is equal to the limit as 𝑥 approaches 𝜋 by two from the right of 𝑓 of 𝑥. And of course, showing that both of these limits exist. Let’s start by finding the limit as 𝑥 approaches 𝜋 by two from the left of 𝑓 of 𝑥. Since 𝑥 is approaching 𝜋 by two from the left, we must have our values of 𝑥 are less than 𝜋 by two. And we can see from our piecewise definition of the function 𝑓 of 𝑥, when our values of 𝑥 are less than 𝜋 by two, our function 𝑓 of 𝑥 is exactly equal to eight plus the cos of seven 𝑥.

And if these functions are exactly equal for values of 𝑥 less than 𝜋 by two, their limits as 𝑥 approaches 𝜋 by two from the left will also be equal. So we want to evaluate the limit as 𝑥 approaches 𝜋 by two from the left of eight plus the cos of seven 𝑥. But this is just a constant plus a trigonometric function. We can do this by direct substitution. Substituting 𝑥 is equal to 𝜋 by two, we get eight plus the cos of seven 𝜋 by two. And again, the cos function is periodic about an interval of two 𝜋. So the cos of seven 𝜋 by two is equal to the cos of three 𝜋 by two, which we know is equal to zero. So we’ve shown this limit evaluates to give us eight plus zero, which is equal to eight. So the limit as 𝑥 approached 𝜋 by two from the left of 𝑓 of 𝑥 was equal to eight.

We now need to show that our limit as 𝑥 approaches 𝜋 by two from the right of 𝑓 of 𝑥 is also equal to eight. And we can actually check this in exactly the same way. We’ll take the limit as 𝑥 approaches 𝜋 by two from the right of 𝑓 of 𝑥. This means that our values of 𝑥 are greater than 𝜋 by two. We then see from our piecewise definition of the function 𝑓 of 𝑥, if our values of 𝑥 are greater than 𝜋 by two, our function 𝑓 of 𝑥 is exactly equal to seven plus the sin of five 𝑥. So their limits as 𝑥 approaches 𝜋 by two from the right will be equal.

And now we see we want to evaluate the limit of a constant plus a trigonometric function. We can do this by direct substitution. Substituting 𝑥 is equal to 𝜋 by two, we get seven plus the sin of five 𝜋 by two. And we actually already evaluated this expression. It’s actually equal to 𝑓 evaluated at 𝜋 by two. So we can just say this is equal to eight. So we’ve shown the limit as 𝑥 approaches 𝜋 by two from the right of 𝑓 of 𝑥 is also equal to eight. So both of these limits exist and they’re equal. So our second continuity condition is also true.

Finally, we just need to check our last continuity condition. The limit as 𝑥 approaches 𝜋 by two of 𝑓 of 𝑥 needs to be equal to 𝑓 evaluated at 𝜋 by two. We actually evaluated both of these expressions in the first and second part of our continuity condition. We evaluated 𝑓 at 𝜋 by two in the first part of our continuity condition. We showed that it was equal to eight.

And it’s worth reiterating at this point when we did our second continuity condition and we showed that the limit from the left and the limit from the right of 𝑓 of 𝑥 were both equal. In this case, we showed that both of these were equal to eight. This is actually equivalent to just showing the limit as 𝑥 approaches 𝜋 by two of 𝑓 of 𝑥 is equal to eight. So we’ve already shown that both of these expressions are equal to eight. So our third continuity condition is also true.

Therefore, since we’ve shown all three of our continuity conditions are true for the function 𝑓 of 𝑥 at 𝑥 is equal to 𝜋 by two. We can conclude that the function is continuous at 𝑥 is equal to 𝜋 by two.