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Find the limit of three π₯ squared minus 18π₯ plus 24 all over π₯ squared minus 16 as π₯ approaches four.
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This is a rational function.
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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.
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We can write this fact out slightly more formally.
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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.
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And so the entire statement can be rephrased as a rational function is continuous on its domain.
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Anyway, how does the statement help us with the limit that we want to evaluate?
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The function we have is a rational function.
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So if four is in the domain of this function, to find the limit, we can just directly substitute four into the function.
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Okay, great!
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So how do we know if four is in the domain?
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Well, we try to evaluate π of four and we see if anything goes wrong.
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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.
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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.
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We substitute four for π₯ in the algebraic rule for π of π₯.
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And simplifying, we get the indeterminant form zero over zero.
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So π of four is undefined and four is not in the domain of our function π.
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Okay, what do we do now?
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It looks like directly substituting didnβt work.
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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.
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And if you remember the factor theorem that suggests they both have a factor of π₯ minus four.
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So letβs try to factorize the polynomials in the numerator and denominator of our rational function.
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We can recognize the denominator π₯ squared minus 16 as the difference of two squares.
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And we see the factor of π₯ minus four that we expected from our knowledge of the factor theorem.
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How about the numerator?
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Well, as evaluating the numerator at π₯ equals four gives a zero, we know it has a factor of π₯ minus four.
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The question is, what is the other factor?
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Well, to have three π₯ squared here, the π₯ term of our other factor must be three π₯.
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And to have a constant term of 24 here, the value of π must be 24 divided by negative four, which is negative six.
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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.
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But in the end, we get the same answer.
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We see that thereβs a common factor of π₯ minus four here.
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Weβd like to cancel this common factor of π₯ minus four.
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Of course, this changes the domain of our function.
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These two functions are equal.
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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.
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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.
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What weβre looking for is the limit of π of π₯ as π₯ approaches four.
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This is the same as the limit of π of π₯ as π₯ tends to four.
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Why?
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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.
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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.
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We now focus on finding the limit of π of π₯ as π₯ approaches four.
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π of π₯ is a rational function.
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And so if four is in its domain, then the value of this limit is just π evaluated at π₯ equals four.
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As before the way to see if four is in the domain of π is just to try evaluating π at π₯ equals four.
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Using the definition of π of π₯, we get three times four minus six over four plus four, which is six over eight.
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And that simplifies to three over four or three-fourths.
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Four therefore is in the domain of π.
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And so we can get rid of this question mark of the equal sign.
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The limit of π of π₯ as π₯ approaches four is π of four.
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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.