Video: Finding the Set on Which a Piecewise-Defined Function Involving Trigonometric Ratios Is Continuous

Suppose that 𝑓(π‘₯) = (5 sin (π‘₯ βˆ’ 3))/(π‘₯ βˆ’ 3) if π‘₯ < 3 and 𝑓(π‘₯) =(5π‘₯Β²)/(9) if π‘₯ β‰₯ 3. Find the set on which 𝑓 is continuous.

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Video Transcript

Suppose that 𝑓 of π‘₯ is equal to five sin of π‘₯ minus three over π‘₯ minus three if π‘₯ is less than three and five π‘₯ squared over nine if π‘₯ is greater than or equal to three. Find the set on which 𝑓 is continuous.

Here, we have a piecewise function defined over two intervals. The boundary of our two intervals occurs at the point where π‘₯ equals three. And hence, this is an important point. For our first subfunction, we have a trigonometric expression on the top-half of our quotient and a binomial on the bottom-half. Since we know that trigonometric functions and polynomials are continuous on their domains, and quotients of continuous functions are also continuous on their domains, we conclude that this subfunction is, therefore, also continuous on its domain.

Now, here, we need to be slightly careful since, at values where π‘₯ equals three, this subfunction evaluates to zero over zero, which is an indeterminate form. Lucky for us, 𝑓 of π‘₯ is only defined by this subfunction at values of π‘₯ which are strictly less than three, not where π‘₯ is equal to three. Instead, when π‘₯ is greater than or equal to three, 𝑓 of π‘₯ is defined by five π‘₯ squared over nine.

Again, here, it’s worth noting that this monomial is defined over and has a domain of all real numbers. This means that the domain of our function 𝑓 of π‘₯ is all the real numbers. But we should be careful not to hastily jump to the conclusion that 𝑓 of π‘₯ is also continuous over all the real numbers. Instead, we must still check the criteria for continuity at the boundary between our subfunctions, or when π‘₯ is equal to three.

First, let’s find 𝑓 of three by substituting in to five π‘₯ squared over nine. This value is easy to compute. And we get an answer of five. Next, we need to check that our normal limit exists and is also equal to five. If this is not the case, then we’ll have a discontinuity at π‘₯ equals three. And hence, our function will not be continuous at this point.

To move forward, we first recognise that, either side of π‘₯ equals three, our function is defined by two different subfunctions. To find our left limit, or when π‘₯ approaches from the negative direction, we use our first subfunction. Here, we’ve already shown that a direct substitution of π‘₯ equals three leads us to an indeterminate form of zero over zero, so we’ll need to use a different approach. Instead, we use the rule that the limit, as π‘₯ approaches zero of sin π‘₯ over π‘₯, is equal to one.

Now, our expression isn’t in this form. And so, we perform some manipulations first by taking a factor of five outside of our limit using the constant multiple rule. Next, we perform a 𝑒-substitution. By setting 𝑒 as π‘₯ minus three, we get the following. Our limit becomes sin 𝑒 over 𝑒. However, we mustn’t forget to change the limit value itself. 𝑒 plus three is equal to π‘₯. Hence, 𝑒 plus three approaches three from the negative direction, or 𝑒 approaches zero from the negative direction.

Looking at our rule, we know that if the normal limit exists and is equal to one, then the left- and right-sided limits also exist and are also equal to one. We can now use our rule to evaluate that this limit is equal to one. Hence, the left-sided limit, as π‘₯ approaches three of 𝑓 of π‘₯, is equal to five times one, which is, of course, just five. Now, our right-sided limit is far easier to evaluate. For this, since we’re approaching π‘₯ equals three from the right, we take the limit using our other subfunction. Simply, by direct substitution, we find that this limit is equal to five.

Since the left- and right-sided limits both exist and are equal to the same value, we can, therefore, conclude that the limit, as π‘₯ approaches three of 𝑓 of π‘₯, is also equal to five. And earlier, you’ll recall that we found that 𝑓 of three is also equal to five. Since the limit, as π‘₯ approaches three of 𝑓 of π‘₯, is equal to 𝑓 of three, we conclude that 𝑓 of π‘₯ is continuous when π‘₯ is equal to three.

If you were to think about this visually, this means that the two end points of our subfunctions will join up. Let us now recall that, earlier, we concluded that 𝑓 of π‘₯ was continuous over all of the real numbers aside from three, which we had to check. And now that we have checked three, we’re in a position to say that our function 𝑓 of π‘₯ is continuous on all the real numbers. And this is the answer to our question.

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