### 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?

Letβs clear some space to discuss
this. 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. Then the limit that we have to find
is the limit of π 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 π of nine.

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. Just as an aside, you might know
that this is a definition for π to be continuous at π₯ equals nine. Donβt worry if you havenβt 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 weβve 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 shouldnβt 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 π is just π for any number π. And the limit of a constant
function π as π₯ approaches π again for any number π is just the constant π.

You may feel that this is cheating,
that these two arenβt really limit laws; theyβre just values of limits. 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.

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. 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. 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.

These laws of limits arenβt
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.