### Video Transcript

Properties of Limits

In this video, we will learn how to
use the properties of limits such as the limits of sums, differences, products, and
quotients of functions and the limits of certain composite functions. Weβll be looking at various
examples of how we can use these properties. Letβ²s start by defining some
properties.

Suppose that π of π₯ and π of π₯
are functions and π is some value such that the limit as π₯ tends to π of π of π₯
and the limit as π₯ tends to π of π of π₯ both exist. Then, we have the property for
limits of sums of functions, which tells us that the limit as π₯ tends to π of π
of π₯ plus π of π₯ is equal to the limit as π₯ tends to π of π of π₯ plus the
limit as π₯ tends to π of π of π₯. We also have a property for the
limits of differences of functions. And this tells us that the limit as
π₯ tends to π of π of π₯ minus π of π₯ is equal to the limit as π₯ tends to π of
π of π₯ minus the limit as π₯ tends to π of π of π₯. And we can note that these two
properties can be used in combination with one another. Letβ²s now look at an example of how
these properties can be used.

Given that the limit as π₯ tends to
two of π of π₯ is equal to three, the limit as π₯ tends to two of π of π₯ is equal
to negative seven, and the limit as π₯ tends to two of β of π₯ is equal to negative
one, find the limit as π₯ tends to two of π of π₯ plus π of π₯ minus β of π₯.

In order to find this limit, we can
start by breaking it down using the properties of limits. We have the property for the limits
of sums of functions which tells us that the limit as π₯ tends to π of π of π₯
plus π of π₯ is equal to the limit as π₯ tends to π of π of π₯ plus the limit as
π₯ tends to π of π of π₯. We also have the property for the
limits of differences of functions. And this tells us that the limit as
π₯ tends to π of π of π₯ minus π of π₯ is equal to the limit as π₯ tends to π of
π of π₯ minus the limit as π₯ tends to π of π of π₯. Now, we can apply these two
properties to the limit weβre trying to find.

We can start by using the rule for
limits of sums of functions. In our case, π is equal to
two. And we can split the inside of our
limit up so that we are adding two functions, these functions being π of π₯ and π
of π₯ minus β of π₯. We obtain that our limit is equal
to the limit as π₯ tends to two of π of π₯ plus the limit as π₯ tends to two of π
of π₯ minus β of π₯. Next, we can use the rule for the
limits of differences of functions. Again, π is equal to two. And inside our limit, we have a
difference of two functions that being π of π₯ and β of π₯. Applying the rule, we obtain that
our limit is equal to the limit as π₯ tends to two of π of π₯ plus the limit as π₯
tends to two of π of π₯ minus the limit as π₯ tends to two of β of π₯.

Now we can spot that we know the
value of each of these three limits since they have been given to us in the
question. So substituting in three, negative
seven, and negative one, we obtain that our limit is equal to three plus negative
seven minus negative one. Simplifying this, we obtain a
solution that the limit as π₯ tends to two of π of π₯ plus π of π₯ minus β of π₯
is equal to negative three.

Now letβ²s cover some more
properties of limits.

Again, we have functions π of π₯
and π of π₯ with some constant values π and π such that the limit as π₯ tends to
π of π of π₯ and the limit as π₯ tends to π of π of π₯ both exist. This time, we have an extra
constant π. And weβ²ll see why this is here in
our first property. This first property is about
multiplicative constants within a limit. It tells us that the limit as π₯
tends to π of π multiplied by π of π₯ is equal to π multiplied by the limit as
π₯ tends to π of π of π₯. Essentially, what this tells us is
that if we have a constant factor within our limit, we can simply factor it outside
of the limit. The next property is the limit of
product of functions. It tells us that the limit as π₯
tends to π of π of π₯ multiplied by π of π₯ is equal to the limit as π₯ tends to
π of π of π₯ multiplied by the limit as π₯ tends to π of π of π₯. Our third property here is for the
limits of quotients of functions. It tells us that the limit as π₯
tends to π of π of π₯ over π of π₯ is equal to the limit as π₯ tends to π of π
of π₯ over the limit as π₯ tends to π of π of π₯. And itβ²s again important to note
that each of these rules can be used in combination with one another, including the
two rules we covered previously. Letβ²s now look at some examples of
how these rules can be used.

Assume that the limit as π₯ tends
to three of π of π₯ is equal to five, the limit as π₯ tends to three of π of π₯ is
equal to eight, and the limit as π₯ tends to three of β of π₯ is equal to nine. Find the limit as π₯ tends to three
of π of π₯ multiplied by π of π₯ minus β of π₯.

We can start by breaking down the
limit given in the question using the properties of limits. Firstly, we can use the rule for
the limits of differences of functions. This tells us that the limit as π₯
tends to π of π of π₯ minus π of π₯ is equal to the limit as π₯ tends to π of π
of π₯ minus the limit as π₯ tends to π of π of π₯. Looking at our limit, we can see
that our value of π is three. And we can see that we have a
difference of functions within our limit. We have π of π₯ multiplied by π
of π₯ minus β of π₯. So we can apply our rule, giving us
that our limit is equal to the limit as π₯ tends to three of π of π₯ times π of π₯
minus the limit as π₯ tends to three of β of π₯.

In order to split down this limit
further, weβll need to use another limit property. And this is the property of limits
of products of functions, which tells us that the limit as π₯ tends to some constant
π of a product of functions β so π of π₯ times π of π₯ β is equal to the limit as
π₯ tends to π of π of π₯ times the limit as π₯ tends to π of π of π₯. Again, our value of π is
three. And we can see that inside our
limit, we have a product of functions. So thatβs π of π₯ times π of
π₯. Applying this property, we find our
limit is equal to the limit as π₯ tends to three of π of π₯ multiplied by the limit
as π₯ tends to three of π of π₯ minus the limit as π₯ tends to three of β of
π₯.

And we can spot that weβ²ve been
given each of these three limits within the question. We have that the limit as π₯ tends
to three of π of π₯ is equal to five, the limit as π₯ tends to three of π of π₯ is
equal to eight, and the limit as π₯ tends to three of β of π₯ is equal to nine. So we substitute these values in
for our limit, giving us that our limit is equal to five timesed by eight minus
nine. This can be simplified to obtain a
solution that the limit as π₯ tends to three of π of π₯ multiplied by π of π₯
minus β of π₯ is equal to 31.

In the next example, we will see
how the property for quotients of functions can be used.

Given that the limit as π₯ tends to
negative two of π of π₯ over three π₯ squared is equal to negative three, determine
the limit as π₯ tends to negative two of π of π₯.

In this question, weβ²ve been given
the limit as π₯ tends to negative two of π of π₯ over three π₯ squared. We can break this limit down using
the properties of limits. We have the property for limits of
quotients of functions, which tells us that the limit as π₯ tends to π of π of π₯
over π of π₯ is equal to the limit as π₯ tends to π of π of π₯ over the limit as
π₯ tends to π of π of π₯. For our limit, weβre taking the
limit as π₯ tends to negative two. Therefore, π is equal to negative
two. And we have a quotient of
functions. In the numerator, we have π of
π₯. And in the denominator, we have
three π₯ squared. Applying this rule for limits of
quotients of functions, we obtain that our limit is equal to the limit as π₯ tends
to negative two of π of π₯ over the limit as π₯ tends to negative two of three π₯
squared.

Letβ²s now consider the limit in the
denominator of the fraction. Thatβ²s the limit as π₯ tends to
negative two of three π₯ squared. We can apply direct substitution to
this limit, giving us that the limit as π₯ tends to negative two of three π₯ squared
is equal to three times negative two squared. Negative two squared is equal to
four. We then simplify to obtain that
this limit is equal to 12. We can substitute this value of 12
back in to the denominator of our fraction, giving us that the limit as π₯ tends to
negative two of π of π₯ over three π₯ squared is equal to the limit as π₯ tends to
negative two of π of π₯ all over 12.

However, weβ²ve been given in the
question that the limit as π₯ tends to negative two of π of π₯ over three π₯
squared is equal to negative three. And since this is on the left-hand
side of our equation, we can set our equation equal to negative three. So we now have that the limit as π₯
tends to negative two of π of π₯ over 12 is equal to negative three. We simply multiply both sides of
the equation by 12. Here, we reach our solution which
is that the limit as π₯ tends to negative two of π of π₯ is equal to negative
36.

There are a couple more limit
properties weβll be covering in this video and these are as follows.

Given some function π of π₯ with
some value π and an integer π such that the limit as π₯ tends to π of π of π₯
exists.

We have the property for the limits
of powers of functions, which tells us that the limit as π₯ tends to π of π of π₯
to the power of π is equal to the limit as π₯ tends to π of π of π₯ all to the
power of π. So essentially, if we have a power
of a function within a limit, we can take the power outside of the limit and simply
raise the whole limit to that power. Letβ²s quickly note that our integer
value of π here can be both positive or negative. And this rule will still work. We can see how this limit property
can be derived from the limits of products of functions and the limits of quotients
of functions properties. Since if we repeated either of
those properties over and over again π times with just one function π of π₯, then
we would obtain this property.

Our final limit property here is
the property for the limits of roots of functions. It tells us that the limit as π₯
tends to π of the πth root of π of π₯ is equal to the πth root of the limit as
π₯ tends to π of π of π₯. So essentially, if we have an πth
root of a function within our limit, then we can take the πth root outside of the
limit and instead take the πth root of limit of the function. These two properties can again be
combined with one another and with any of the previous properties. Letβ²s look at some examples of how
they can be used.

Assume that the limit as π₯ tends
to six of π of π₯ is equal to three and the limit as π₯ tends to six of π of π₯ is
equal to eight. Find the limit as π₯ tends to six
of the square root of π of π₯ minus π of π₯.

We need to find the limit as π₯
tends to six of the square root of π of π₯ minus π of π₯. We can break this limit down using
the properties of limits. We have the property for the limits
of roots of functions. It tells us that the limit as π₯
tends to some constant π of the πth root of some function π of π₯ is equal to the
πth root of the limit as π₯ tends to π of π of π₯. The limit weβre trying to find is
the limit as π₯ tends to six of the square root of π of π₯ minus π of π₯. So we have the limit of a square
root of a function. We can therefore apply our rule for
limits of roots of functions. It tells us that our limit is equal
to the square root of the limit as π₯ tends to six of π of π₯ minus π of π₯.

Now, we can see that we have the
limit of a difference of functions since our limit is of π of π₯ minus π of
π₯. We can apply the rule for the limit
of differences of functions, which tells us that the limit as π₯ tends to some
constant π of a difference of functions β so thatβs π of π₯ minus π of π₯ β is
equal to the limit as π₯ tends to π of π of π₯ minus the limit as π₯ tends to π
of π of π₯. We can apply this rule to our limit
within the square root, giving us that our limit is equal to the square root of the
limit as π₯ tends to six of π of π₯ minus the limit as π₯ tends to six of π of
π₯.

Now, we can spot that the limits
within our square root have been given to us in the question. We have that the limit as π₯ tends
to six of π of π₯ is equal to three and the limit as π₯ tends to six of π of π₯ is
equal to eight, giving us that our limit is equal to the square root of eight minus
three. Simplifying this, we obtain our
solution which is that the limit as π₯ tends to six of the square root of π of π₯
minus π of π₯ is equal to the square root of five.

Next, weβ²ll move onto our final
example, where we will see how the properties of limits can be used even when the
function is defined graphically.

Consider the graph of π of π₯. Find the limit as π₯ tends to one
of π₯ multiplied by π of π₯ all squared.

Here, weβve been asked to find the
limit as π₯ tends to one of π₯ multiplied by π of π₯ all squared. We can see that we have a power of
a function. And so, we can use our rule for the
limits of powers of functions. It tells us that the limit as π₯
tends to π of π of π₯ to the power of π is equal to the limit as π₯ tends to π
of π of π₯ all to the power of π. In the case of our limit, the value
of π is one and the power which we are raising our function to is two. So π is equal to two. Now, we can apply this rule. It tells us that our limit is equal
to the limit as π₯ tends to one of π₯ multiplied by π of π₯ all squared.

Next, we can in fact simplify the
limit within our square further. We have the rule for the limits of
product of functions. It tells us that the limit as π₯
tends to π of π of π₯ multiplied by π of π₯ is equal to the limit as π₯ tends to
π of π of π₯ multiplied by the limit as π₯ tends to π of π of π₯. Now our product of functions is π₯
multiplied by π of π₯. So when we apply this rule to our
limit, we obtain the limit as π₯ tends to one of π₯ multiplied by the limit as π₯
tends to one of π of π₯ all squared.

Now, letβ²s consider the limit as π₯
tends to one of π₯. We can apply direct substitution to
this limit. And we obtain that it is equal to
one. So we can substitute this back in
for our limit, giving us that the limit as π₯ tends to one of π₯ timesed by π of π₯
squared is equal to the limit as π₯ tends to one of π of π₯ all squared. Now all we need to do is find the
limit as π₯ tends to one of π of π₯. In order to do this, we need to use
our graph. We need to find the value of π of
π₯ when π₯ is equal to one. We see that when π₯ is equal to
one, π of π₯ is equal to three. And the graph of π of π₯ near to
the π₯-value of one is a straight line. Therefore, the right limit of π of
π₯ and the left limit of π of π₯ will both agree that this limit is equal to
three. So we can substitute in three for
the limit as π₯ tends to one of π of π₯. So we find that our limit is equal
to three squared. And we can square the three to
obtain our solution that the limit as π₯ tends to one of π₯ multiplied by π of π₯
squared is equal to nine.

Now we have covered a variety of
examples letβ²s recap some key points of the video.

Key Points

For functions π of π₯ and π of π₯
with values π and π and integer π such that the limit as π₯ tends to π of π of
π₯ and the limit as π₯ tends to π of π of π₯ exist: the limit as π₯ tends to π of
π of π₯ plus π of π₯ is equal to the limit as π₯ tends to π of π of π₯ plus the
limit as π₯ tends to π of π of π₯. The limit as π₯ tends to π of π
of π₯ minus π of π₯ is equal to the limit as π₯ tends to π of π of π₯ minus the
limit as π₯ tends to π of π of π₯. The limit as π₯ tends to π of π
multiplied by π of π₯ is equal to π multiplied by the limit as π₯ tends to π of
π of π₯. The limit as π₯ tends to π of π
of π₯ multiplied by π of π₯ is equal to the limit as π₯ tends to π of π of π₯
multiplied by the limit as π₯ tends to π of π of π₯. The limit as π₯ tends to π of π
of π₯ over π of π₯ is equal to the limit as π₯ tends to π of π of π₯ over the
limit as π₯ tends to π of π of π₯. The limit as π₯ tends to π of π
of π₯ to the power of π is equal to the limit as π₯ tends to π of π of π₯ all to
the power of π. The limit as π₯ tends to π of the
πth root of π of π₯ is equal to the πth root of the limit as π₯ tends to π of π
of π₯. And all of these limit properties
can be used in combination with one another.