Video: Finding the Limit of Root Functions at a Point by Direct Substitution

Determine lim_(π‘₯ β†’ 9) √(4π‘₯Β² βˆ’ 9π‘₯ + 1).

04:58

Video Transcript

Determine the limit of the square root of four π‘₯ squared minus nine π‘₯ plus one as π‘₯ approaches nine.

It turns out that we can evaluate this limit by direct substitution. The value that π‘₯ is approaching is nine. And we just substitute this value for π‘₯ in the expression we have.

We now have the square root of four times nine squared minus nine times nine plus one. Now all we have to do is to evaluate this radical. Underneath the radical symbol, we get 244. And as 244 is two squared times 61, we can simplify this radical. And doing so, we get two square root 61. This is the value of the limit that we get by directly substituting. But what justifies us using direct substitution to evaluate this limit?

Let’s clear some space to discuss this. Let 𝑓 of π‘₯ be the thing that we’re trying to find the limit of, that is, the square root of four π‘₯ squared minus nine π‘₯ plus one. Then the limit that we have to find is the limit of 𝑓 of π‘₯ as π‘₯ approaches nine. 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 𝑓 of nine.

And here you can check that the square root of four times nine squared minus nine times nine plus one is really 𝑓 of nine. Just as an aside, you might know that this is a definition for 𝑓 to be continuous at π‘₯ equals nine. Don’t worry if you haven’t come across this definition. The important thing is that we can justify our use of direct substitution using the laws of limits.

The first law that we use is that we can swap the order in which we take the limit and the square root. 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.

And you might know already that you can evaluate the limit inside the square root by using direct substitution. And doing so would take us to the line above and from there to our answer.

Alternatively, we can use another limit law: that the limit of a sum of functions is the sum of the limits of the functions. And this actually holds no matter how many functions there are in the sum. So we can find the limit of each term individually. These three limits can be evaluated by direct substitution. And doing so would take us to the top line and our answer.

But as we’ve come this far, we might as well justify why we can use direct substitution to find these limits. The limit of a number times a function is just that number times the limit of the function. So the limit of four times π‘₯ squared is four times the limit of π‘₯ squared. And the limit of nine times π‘₯ is nine times the limit of π‘₯. We leave the limit of one as it is.

One more thing we can do is to write the limit of π‘₯ squared as the square of the limit of π‘₯. 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.

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. 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. But shouldn’t we justify the use of direct substitution even to such simple functions?

Well, actually, we turn these into laws. The limit of the identity function π‘₯ as π‘₯ approaches 𝑐 is just 𝑐 for any number 𝑐. And the limit of a constant function π‘˜ as π‘₯ approaches 𝑐 again for any number 𝑐 is just the constant π‘˜.

You may feel that this is cheating, that these two aren’t really limit laws; they’re just values of limits. But the truth is if we can only write a limit in terms of some other limits, then we’re never going to be able to evaluate any of them.

The laws of limits, including the limit of the identity function 𝑓 of π‘₯ equals π‘₯ and the limit of a constant function 𝑓 of π‘₯ equals π‘˜, are the ingredients we need to find other limits. If you agree with the laws of limits as we’ve written them, then you’ll 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. And if you don’t agree that this is a value of our limit, then you should be able to point out which limit law you disagree with.

These laws of limits aren’t arbitrary either. They should seem relatively reasonable if you think about what a limit means intuitively. And these limit laws can also be proved using a more formal definition of limit involving epsilons and deltas, which you might see later.

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