### Video Transcript

Find the limit of three π₯ squared minus 18π₯ plus 24 all over π₯ squared minus 16 as π₯ approaches four.

This is a rational function. And we know that we can evaluate the limit of a rational function as π₯ approaches some value in this domain by just directly substituting. We can write this fact out slightly more formally. Just as an aside, when the limit of π of π₯ as π₯ approaches π is just equal to π of π, we say that π of π₯ is continuous at π. And so the entire statement can be rephrased as a rational function is continuous on its domain.

Anyway, how does the statement help us with the limit that we want to evaluate? The function we have is a rational function. So if four is in the domain of this function, to find the limit, we can just directly substitute four into the function.

Okay, great! So how do we know if four is in the domain? Well, we try to evaluate π of four and we see if anything goes wrong. If four is in the domain, then π of four will be defined and its value will be the value of the limits that weβre looking for. If four is not in the domain, then weβll find that π of four is not defined and weβll have to use a different method to evaluate the limit.

We substitute four for π₯ in the algebraic rule for π of π₯. And simplifying, we get the indeterminant form zero over zero. So π of four is undefined and four is not in the domain of our function π. Okay, what do we do now? It looks like directly substituting didnβt work. We got the value zero over zero when we try to find π of four because the polynomials in the numerator and denominator of our function are both zero when π₯ is four. And if you remember the factor theorem that suggests they both have a factor of π₯ minus four.

So letβs try to factorize the polynomials in the numerator and denominator of our rational function. We can recognize the denominator π₯ squared minus 16 as the difference of two squares. And we see the factor of π₯ minus four that we expected from our knowledge of the factor theorem.

How about the numerator? Well, as evaluating the numerator at π₯ equals four gives a zero, we know it has a factor of π₯ minus four. The question is, what is the other factor? Well, to have three π₯ squared here, the π₯ term of our other factor must be three π₯. And to have a constant term of 24 here, the value of π must be 24 divided by negative four, which is negative six.

Of course, there are other methods we could have used to factor this numerator, for example, taking the common factor of three out of the coefficients. But in the end, we get the same answer. We see that thereβs a common factor of π₯ minus four here. Weβd like to cancel this common factor of π₯ minus four.

Of course, this changes the domain of our function. These two functions are equal. That is they give the same output for a given input apart from four π₯ equals four when the function three π₯ squared minus 18π₯ plus 24 all over π₯ squared minus 16 is undefined.

Another way of saying this is if we let π of π₯ be three π₯ minus six over π₯ plus four β the simplified version of π of π₯ in some sense β then π of π₯ is π of π₯ if π₯ is not equal to four and undefined if π₯ equals four.

What weβre looking for is the limit of π of π₯ as π₯ approaches four. This is the same as the limit of π of π₯ as π₯ tends to four. Why? Well, because the limit of π of π₯ as π₯ approaches four depends on values of π₯ near four, but importantly not π₯ equals four. And for all values of π₯ and so in particular all values of π₯ near four, π of π₯ and π of π₯ are identical.

We now focus on finding the limit of π of π₯ as π₯ approaches four. π of π₯ is a rational function. And so if four is in its domain, then the value of this limit is just π evaluated at π₯ equals four. As before the way to see if four is in the domain of π is just to try evaluating π at π₯ equals four. Using the definition of π of π₯, we get three times four minus six over four plus four, which is six over eight. And that simplifies to three over four or three-fourths.

Four therefore is in the domain of π. And so we can get rid of this question mark of the equal sign. The limit of π of π₯ as π₯ approaches four is π of four. And so following the chain of equal signs and remembering how we defined π of π₯, we see that the limit of three π₯ squared minus 18π₯ plus 24 all over π₯ squared minus 16 as π₯ approaches four is three- fourths.