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Determine the limit of the square root of four π₯ squared minus nine π₯ plus one as π₯ approaches nine.
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It turns out that we can evaluate this limit by direct substitution.
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The value that π₯ is approaching is nine.
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And we just substitute this value for π₯ in the expression we have.
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We now have the square root of four times nine squared minus nine times nine plus one.
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Now all we have to do is to evaluate this radical.
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Underneath the radical symbol, we get 244.
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And as 244 is two squared times 61, we can simplify this radical.
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And doing so, we get two square root 61.
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This is the value of the limit that we get by directly substituting.
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But what justifies us using direct substitution to evaluate this limit?
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Letβs clear some space to discuss this.
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Let π of π₯ be the thing that weβre trying to find the limit of, that is, the square root of four π₯ squared minus nine π₯ plus one.
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Then the limit that we have to find is the limit of π of π₯ as π₯ approaches nine.
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And we assume that we could find this limit by just directly substituting nine for π₯, giving us on the right-hand side of the equation just π of nine.
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And here you can check that the square root of four times nine squared minus nine times nine plus one is really π of nine.
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Just as an aside, you might know that this is a definition for π to be continuous at π₯ equals nine.
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Donβt worry if you havenβt come across this definition.
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The important thing is that we can justify our use of direct substitution using the laws of limits.
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The first law that we use is that we can swap the order in which we take the limit and the square root.
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So the limit of the square root of four π₯ squared minus nine π₯ plus one is just the square root of the limit of four π₯ squared minus nine π₯ plus one.
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And you might know already that you can evaluate the limit inside the square root by using direct substitution.
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And doing so would take us to the line above and from there to our answer.
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Alternatively, we can use another limit law: that the limit of a sum of functions is the sum of the limits of the functions.
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And this actually holds no matter how many functions there are in the sum.
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So we can find the limit of each term individually.
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These three limits can be evaluated by direct substitution.
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And doing so would take us to the top line and our answer.
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But as weβve come this far, we might as well justify why we can use direct substitution to find these limits.
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The limit of a number times a function is just that number times the limit of the function.
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So the limit of four times π₯ squared is four times the limit of π₯ squared.
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And the limit of nine times π₯ is nine times the limit of π₯.
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We leave the limit of one as it is.
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One more thing we can do is to write the limit of π₯ squared as the square of the limit of π₯.
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This comes from yet another of our limit laws: that the limit of a power of a function is that power of the limit of the function.
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Now we have the value that we want in terms of the limit of π₯ as π₯ approaches nine and the limit of the constant function one as π₯ approaches nine.
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And if we accept that the values of these limits are nine and one, respectively, then we get what we have in the top line.
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But shouldnβt we justify the use of direct substitution even to such simple functions?
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Well, actually, we turn these into laws.
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The limit of the identity function π₯ as π₯ approaches π is just π for any number π.
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And the limit of a constant function π as π₯ approaches π again for any number π is just the constant π.
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You may feel that this is cheating, that these two arenβt really limit laws; theyβre just values of limits.
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But the truth is if we can only write a limit in terms of some other limits, then weβre never going to be able to evaluate any of them.
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The laws of limits, including the limit of the identity function π of π₯ equals π₯ and the limit of a constant function π of π₯ equals π, are the ingredients we need to find other limits.
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If you agree with the laws of limits as weβve written them, then youβll have to agree that the limit of the square root of four π₯ squared minus nine π₯ plus one as π₯ approaches nine is two root 61.
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And if you donβt agree that this is a value of our limit, then you should be able to point out which limit law you disagree with.
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These laws of limits arenβt arbitrary either.
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They should seem relatively reasonable if you think about what a limit means intuitively.
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And these limit laws can also be proved using a more formal definition of limit involving epsilons and deltas, which you might see later.