Video: Finding an Unknown in a Piecewise-Defined Function Involving Trigonometric Ratios That Makes It Continuous at a Point

Find the value of π‘˜ which makes the function 𝑓 continuous at π‘₯ = 0, given that 𝑓(π‘₯) = (sin 2π‘₯ tan 4π‘₯)/(7π‘₯Β²) if π‘₯ β‰  0 and 𝑓(π‘₯) = 7π‘˜ if π‘₯ = 0.

06:12

Video Transcript

Find the value of π‘˜ which makes the function 𝑓 continuous at π‘₯ is equal to zero, given that 𝑓 of π‘₯ is equal to the sin of two π‘₯ tan of four π‘₯ divided by seven π‘₯ squared if π‘₯ is not equal to zero, and 𝑓 of π‘₯ is equal to seven π‘˜ if π‘₯ is equal to zero.

The question wants us to find the value of π‘˜ which will make our piecewise function 𝑓 continuous at the point where π‘₯ is equal to zero. And we recall that we call a function 𝑓 continuous at the point π‘₯ is equal to π‘Ž if the following three conditions are true. First, the function must be defined at the point where π‘₯ is equal to π‘Ž. And it’s worth noting, this is equivalent to saying that π‘Ž is in the domain of our function 𝑓 of π‘₯.

Second, the limit as π‘₯ approaches π‘Ž of our function 𝑓 of π‘₯ must exist. And 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. Thirdly, we must have the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ is equal to 𝑓 evaluated at π‘Ž. Since the question wants us to find the value of π‘˜ which will make our function 𝑓 continuous when π‘₯ is equal to zero, we’ll set π‘Ž equal to zero in our definition of continuity. And we’ll need to find the value of π‘˜ which makes all three of our conditions true.

First, we need to check that 𝑓 of π‘₯ is defined when π‘₯ is equal to zero. And we see from our piecewise definition of the function 𝑓 of π‘₯, when π‘₯ is equal to zero, our function outputs seven π‘˜. So, 𝑓 evaluated at zero is equal to seven π‘˜. This means that our function is defined when π‘₯ is equal to zero. Next, we need to check the limit as π‘₯ approaches zero of our function 𝑓 of π‘₯ exists. We’ll do this by checking 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‘₯.

We’ll start by checking the limit as π‘₯ approaches zero from the right of our function 𝑓 of π‘₯. Since π‘₯ is approaching zero from the right, we must have that π‘₯ is greater than zero. And if π‘₯ is greater than zero, that means π‘₯ is not equal to zero. And we see that our function 𝑓 of π‘₯ is equal to the sin of two π‘₯ times tan of four π‘₯ all divided by seven π‘₯ squared when π‘₯ is not equal to zero. So, when π‘₯ is not equal to zero, both of these functions are exactly the same. That means their limits as π‘₯ approaches zero from the right are equal. We might be tempted at this point to try direct substitution. However, we’ll get a numerator of the sin of zero multiplied by the tan of zero, which is zero, and a denominator of seven multiplied by zero squared, giving us the indeterminate form zero divided by zero.

So, we’ll need to perform some kind of manipulation to evaluate this limit. We’ll start by using the fact that the tan of four π‘₯ is equivalent to the sin of four π‘₯ divided by the cos of four π‘₯. Replacing the tan of four π‘₯ with the sin of four π‘₯ over the cos of four π‘₯ gives us the limit as π‘₯ approaches zero from the right of the sin of two π‘₯ sin of four π‘₯ all divided by seven π‘₯ squared cos of four π‘₯. However, we still can’t use direct substitution to evaluate this limit. Since π‘₯ is approaching zero, we’ll get a factor of the sin of zero, which is zero, in our numerator and a factor of zero in our denominator.

We’re going to use one of our standard trigonometric limit results. For any constant π‘Ž, the limit as π‘₯ approaches zero of the sin of π‘Žπ‘₯ divided by π‘₯ is equal to π‘Ž. This is a really useful limit result which we should commit to memory. To use this result to evaluate our limit, we’re going to need to rewrite our limit. We’ll take out the sin of two π‘₯ in our numerator and one of the π‘₯s in our denominator for our first factor. Then, we’ll take out the sin of four π‘₯ in our numerator and the other factor of π‘₯ in our denominator for our second factor. This leaves us with one divided by seven cos of four π‘₯.

Since we know the limit of a product is equal to the product of a limit, we can evaluate each limit of our factor separately. This means we’re now ready to evaluate our limit. Using the limit as π‘₯ approaches zero of the sin of π‘Žπ‘₯ over π‘₯ is equal to π‘Ž, the limit of our first factor is two, and the limit of our second factor is four. And we can evaluate the limit of our third factor using direct substitution since it’s made up of standard trigonometric functions. So, we substitute π‘₯ is equal to zero. This means that our limit was equal to eight divided by seven times cos of zero, which simplifies to give us eight over seven.

So, we’ve shown the limit as π‘₯ approaches zero from the right of 𝑓 of π‘₯ is equal to eight over seven. We now need to check the limit as π‘₯ approaches zero from the left of 𝑓 of π‘₯ exists and is equal. We’ll do this by asking the question, what would’ve happened to our working if instead we’d had the limit as π‘₯ approaches zero from the left? Now, instead of concluding that π‘₯ is greater than zero, since π‘₯ is approaching zero from the left, we have that π‘₯ is less than zero. However, we only use the fact that π‘₯ was not equal to zero. So in fact, our next step of working is exactly the same. In fact, none of our lines of working specifically used the fact that π‘₯ is approaching zero from the right.

So, all of our lines of working will be the same and we’ll show the limit as π‘₯ approaches zero from the left of 𝑓 of π‘₯ is also equal to eight over seven. Therefore, since we’ve shown 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‘₯ both exist and are equal, our second continuity condition is also true. Finally, for 𝑓 to be continuous at π‘₯ is equal to zero, we need to show the limit as π‘₯ approaches zero of 𝑓 of π‘₯ is equal to 𝑓 evaluated at zero.

We’ve already shown that 𝑓 evaluated at zero is equal to seven π‘˜. And when verifying the second continuity condition, we showed that the limit as π‘₯ approaches zero from the left of 𝑓 of π‘₯ and the limit as π‘₯ approaches zero from the right of 𝑓 of π‘₯ will both be equal to eight over seven. By definition, this tells us that the limit as π‘₯ approaches zero of 𝑓 of π‘₯ is equal to eight over seven. So, for our third continuity condition to be true, we must have that seven π‘˜ is equal to eight divided by seven. Dividing both sides of this equation by seven gives us that π‘˜ is equal to eight divided by 49.

Therefore, we’ve shown for the function 𝑓 of π‘₯ is equal to sin two π‘₯ tan four π‘₯ over seven π‘₯ squared when π‘₯ is not equal to zero and 𝑓 of π‘₯ is equal to seven π‘˜ when π‘₯ is equal to zero. To be continuous at the point where π‘₯ is equal to zero, we need the value of π‘˜ to be eight divided by 49.

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