Video: Finding Unknowns in a Piecewise-Defined Function Involving Trigonometric Ratios That Make It Continuous at a Point

Find the value of π‘˜ which makes the function 𝑓 continuous at π‘₯ = 0, given 𝑓(π‘₯) = sin 2π‘₯ cot 3π‘₯ if π‘₯ β‰  0 and 𝑓(π‘₯) = π‘˜ if π‘₯ = 0.

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Video Transcript

Find the value of π‘˜ which makes the function 𝑓 continuous at the point where π‘₯ is equal to zero, given that 𝑓 of π‘₯ is equal to the sin of two π‘₯ multiplied by the cot of three π‘₯ if π‘₯ is not equal to zero and 𝑓 of π‘₯ is equal to π‘˜ if π‘₯ is equal to zero.

The question is asking us to find the value of π‘˜ which will make our function 𝑓 continuous at the point where π‘₯ is equal to zero. We’re given a piecewise definition of our function 𝑓 of π‘₯. It’s equal to sin two π‘₯ cot three π‘₯ if π‘₯ is not equal to zero and is equal to π‘˜ when π‘₯ is equal to zero. Since we want to find the value of π‘˜ which will make our function 𝑓 continuous when π‘₯ is equal to zero, we’ll recall the definition of continuity at the point π‘₯ is equal to π‘Ž.

We call a function 𝑓 continuous at the point π‘₯ is equal to π‘Ž if it satisfies the following three conditions. The function 𝑓 must be defined when π‘₯ is equal to π‘Ž. This is the same as saying that π‘Ž is in the domain of our function 𝑓. Second, we must have the limit as π‘₯ approaches π‘Ž of our function 𝑓 of π‘₯ exists. Another way of saying this is that 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‘₯ are both equal. And in particular, both our left-hand and right-hand limit must exist. Finally, we need the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ is equal to 𝑓 evaluated at π‘Ž.

We’re looking at the continuity of our function 𝑓 at the point where π‘₯ is equal to zero. So, we’ll set π‘Ž equal to zero and 𝑓 equal to 𝑓 of π‘₯ in our definition of continuity. Since the question wants us to make the function 𝑓 of π‘₯ continuous at π‘₯ is equal to zero, all three parts of our definition of continuity must be true. So, we’ll check all three parts individually. First, our function 𝑓 of π‘₯ must be defined when π‘₯ is equal to zero. We see that 𝑓 of π‘₯ is defined piecewise. And we’re told that when π‘₯ is equal to zero, 𝑓 of π‘₯ is equal to π‘˜. So, we have that 𝑓 evaluated at zero is equal to π‘˜. This gives us that zero is in the domain of our function 𝑓 of π‘₯. So, the first part of our definition of continuity is true.

Next, since 𝑓 of π‘₯ is continuous, the limit as π‘₯ approaches zero of 𝑓 of π‘₯ must exist. And we know this is the same as saying the limit as π‘₯ approaches zero from the right of 𝑓 of π‘₯ and the limit as π‘₯ approaches zero from the left of 𝑓 of π‘₯ are both equal and they both exist. So, since we have to have that both of these limits exist and are equal, let’s check the limit as π‘₯ approaches zero from the right of 𝑓 of π‘₯.

Since π‘₯ is approaching zero from the right, π‘₯ is always bigger than zero. And we notice that in our definition of our function 𝑓 of π‘₯, if π‘₯ is not equal to zero, then 𝑓 of π‘₯ is equal to the function sin of two π‘₯ cot of three π‘₯. So, when we’re taking a limit as π‘₯ approaches zero from the right, π‘₯ is never equal to zero. So, our function 𝑓 of π‘₯ is exactly equal to sin of two π‘₯ cot of three π‘₯. So, these limits will be equal.

At this point, we might want to try direct substitution. However, since π‘₯ is approaching zero, we notice that cot of three π‘₯ does not exist. So, we’ll need to perform some manipulation to help us evaluate this limit. Let’s rewrite this limit in terms entirely of the sine and cosine function. By definition, we have the cot of π‘₯ is equivalent to one divided by the tan of π‘₯, which is again equivalent to the cos of π‘₯ divided by the sin of π‘₯. And since this is true for any value of π‘₯, we can replace π‘₯ with three π‘₯, giving us the cot of three π‘₯ is equivalent to the cos of three π‘₯ divided by the sin of three π‘₯. Substituting this into our limit gives us the limit as π‘₯ approaches zero from the right of sin of two π‘₯ multiplied by cos of three π‘₯ over sin of three π‘₯.

We might be tempted again to try direct substitution at this point. However, since π‘₯ is approaching zero, in our numerator, we have the sin of two times zero, which is zero. And in our denominator, we have the sin of three times zero, which is also zero, giving us an indeterminate form. So, we’re gonna have to perform more manipulation to evaluate this limit. To help us evaluate this limit, we’re going to use one of our standard trigonometric limit results: the limit as π‘₯ approaches zero of the sin of π‘₯ over π‘₯ is equal to one. And since the limit of reciprocal is equal to the reciprocal of the limit, we have the limit as π‘₯ approaches zero of π‘₯ divided by the sin of π‘₯ is equal to the reciprocal of one, which is just equal to one.

To help us evaluate this limit, we’re going to rewrite our limit as the limit as π‘₯ approaches zero from the right of the sin of two π‘₯ over one times one divided by the sin of three π‘₯ times cos of three π‘₯. Finding this limit is the same as finding the limit if we multiplied the numerator and the denominator by π‘₯. And since we know the limit of a product is equal to the product of a limit, we can find the limit of each factor separately.

To evaluate the limit as π‘₯ approaches zero from the right of the sin of two π‘₯ divided by π‘₯, we replace π‘₯ with two π‘₯ in our limit of sin π‘₯ over π‘₯. And if two π‘₯ is approaching zero, then π‘₯ is approaching zero. So, we have the limit as π‘₯ approaches zero of the sin of two π‘₯ over two π‘₯ is equal to one. And we see the factor of one-half is just a constant. So, we can take it outside of our limit. In fact, we can then multiply through by two, giving us the limit as π‘₯ approaches zero of the sin of two π‘₯ over π‘₯ is equal to two.

We can do the same to evaluate the limit as π‘₯ approaches zero from the right of π‘₯ divided by the sin of three π‘₯. We replace π‘₯ with three π‘₯ in our limit law. If three π‘₯ is approaching zero, this is the same as saying π‘₯ is approaching zero. And then, we take the constant factor of three outside of our limit, giving us the limit as π‘₯ approaches zero of π‘₯ divided by the sin of three π‘₯ is equal to one-third. So, using the limit of the product is equal to the product of the limit, we have that our limit is equal to two multiplied by a third multiplied by the limit as π‘₯ approaches zero from the right of the cos of three π‘₯.

And we’re now evaluating the limit of a standard trigonometric function. We can do this by using direct substitution. Substituting π‘₯ is equal to zero gives us the cos of three times zero, which is the cos of zero, which evaluates to give us one. Therefore, we’ve shown the limit as π‘₯ approaches zero from the right of our function 𝑓 of π‘₯ is equal to two-thirds. So, we found the limit as π‘₯ approaches zero from the right of our function 𝑓 of π‘₯.

Since we want our function 𝑓 of π‘₯ to be continuous, we need both the left-hand and right-hand limit to exist and be equal. What would have happened to our line of working if instead of taking the limit as π‘₯ approaches zero from the right, we’d taken the limit as π‘₯ approaches zero from the left? When π‘₯ is approaching zero from the left, we now have that π‘₯ is less than zero. And we still see when π‘₯ is less than zero, π‘₯ is not equal to zero. So, our function 𝑓 of π‘₯ is still equal to the sin of two π‘₯ multiplied by the cot of three π‘₯. And then, we see that none of our steps specifically used the fact that π‘₯ was approaching zero from the right. So, all of our working would remain the same. And we would show the limit as π‘₯ approaches zero from the left of 𝑓 of π‘₯ is also equal to two-thirds. So, in particular, we’ve shown the limit as π‘₯ approaches zero of our function 𝑓 of π‘₯ exists.

Finally, for our function to be continuous, we need the limit of 𝑓 of π‘₯ as π‘₯ approaches zero is equal to 𝑓 evaluated at zero. So, for a function 𝑓 of π‘₯ to be continuous when π‘₯ is equal to zero, we need 𝑓 evaluated at zero is equal to the limit as π‘₯ approaches zero of 𝑓 of π‘₯, which we’ve shown is equal to two-thirds. However, by using the piecewise definition of our function 𝑓 of π‘₯, we know that 𝑓 evaluated at zero is equal to π‘˜. Therefore, since 𝑓 evaluated at zero needs to be equal to two-thirds for our function 𝑓 to be continuous when π‘₯ is equal to zero, we must have that π‘˜ is equal to two-thirds.

Therefore, we’ve shown for the function 𝑓 of π‘₯ is equal to the sin of two π‘₯ cot of three π‘₯ if π‘₯ is not equal to zero and π‘˜ when π‘₯ is equal to zero will only be continuous at the point when π‘₯ is equal to zero when π‘˜ is equal to two-thirds.

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