### Video Transcript

In this video, weβre going to learn
about limits at infinity and unbounded limits. Weβve learnt previously about the
meaning of the limit of π of π₯ as π₯ approaches some real number π. If the value of this limit is πΏ,
this means that if we choose π₯ close enough to π, then we can make π of π₯ as
close to πΏ as weβd like. We can get the value of π of π₯
arbitrarily close to πΏ. In this video, weβre going to
interpret limits of the form the limit of π of π₯ as π₯ approaches infinity. What does it mean for such limits
to have the value πΏ?

We can try just replacing π in the
above definition by infinity. And so, we interpret this to mean
that if we choose π₯ close enough to infinity, then we can make π of π₯ as close to
πΏ as weβd like. But what does it mean for π₯ be
close enough to infinity when infinity is infinitely far away from any value of π₯
that we could choose? It turns out that instead of saying
that π₯ is close enough to infinity, we should say that π₯ is large enough. So, the limit of π of π₯ as π₯
approaches infinity equals πΏ means that if we choose π₯ large enough, then we can
make π of π₯ as close to πΏ as weβd like.

Looking at the graph of the
reciprocal function, we can see that if we choose π₯ large enough, then we can make
the reciprocal function, one over π₯, as close to zero as weβd like. And so, we say that the limit of
one over π₯, as π₯ approaches infinity, is zero. The value of this limit zero is the
value that the function one over π₯ gets closer and closer to as π₯ increases
without bound.

We can also think about the limit
of π of π₯ as π₯ approaches negative infinity. This limit being πΏ means that if
we choose π₯ large and negative enough, in other words π₯ is negative but large
enough in magnitude, then we can make π of π₯ as close to πΏ as weβd like. As with the limit as π₯ approaches
positive infinity, we can think about this value πΏ in a different way. This value πΏ is the value that π
of π₯ gets closer and closer to as π₯ decreases without bound.

So, what is the limit of one over
π₯ as π₯ approaches negative infinity? Well, as π₯ decreases without
bounds, one over π₯ gets closer and closer to zero. So, the value of this limit, again,
is zero. These two limits are very useful
limits to know. It turns out that the limit laws we
learnt for finite limits, suitably interpreted, work just as well for infinite
limits. Using these laws of limits along
with the limits of the reciprocal function as π₯ approaches infinity and negative
infinity that weβve just found, we can find the value of many other limits. Letβs see an example.

Find the limit of negative four
over π₯ squared plus five over π₯ plus eight as π₯ approaches infinity.

We have a limit as π₯ approaches
infinity here, but all the normal rules of limits still apply. For example, the limit of a sum of
functions is equal to the sum of the limits. And so, we can split our limit up
into three. Itβs equal to the limit of negative
four over π₯ squared as π₯ approaches infinity plus the limit of five over π₯ as π₯
approaches infinity plus the limit of eight as π₯ approaches infinity.

What can we say about this
limit? Well, we know that the limit of a
constant πΎ, as π₯ approaches some number π, is just πΎ. And as for the previous limit law,
this holds true, even if π isnβt a real number but is infinity or negative
infinity. The value of this last limit is
just eight.

What about the other two
limits? We can use the fact that the limit
of a constant multiple of a function is that constant multiple of the limit of the
function. The first limit is, therefore,
negative four times the limit of one over π₯ squared as π₯ approaches infinity. And the second is five times the
limit of one over π₯ as π₯ approaches infinity. And finally, we add the eight.

Now, the limit of the reciprocal
function one over π₯, as π₯ approaches infinity, is something we should know. Its value is zero. But how about the limit of one over
π₯ squared as π₯ approaches infinity? Well, we can use the fact that the
limit of a power of a function is that power of the limit of the function. This limit is the limit of the
reciprocal function one over π₯ squared, as one over π₯ squared equals one over π₯
all squared. And by our limit law, this is the
limit of one over π₯ as π₯ approaches infinity all squared. This limit is known to be zero. And so, our limit, the limit of one
over π₯ squared as π₯ approaches infinity, is also zero.

We can generalize in that you get
another limit law that the limit of one over π₯ to the power of π, as π₯ approaches
infinity, is zero, at least if π is greater than zero. Our original limit is, therefore,
negative four times zero plus five times zero plus eight, which is, of course, just
eight. Letβs see another example.

Find the limit of negative two π₯
to the negative four plus eight π₯ to the negative three minus π₯ to the negative
two plus nine π₯ to the negative one minus four all over two π₯ to the negative four
minus six π₯ to the negative three plus seven π₯ to the negative two plus six π₯ to
the negative one plus three, as π₯ approaches infinity.

This is the limit of a quotient of
functions. And we know that the limit of a
quotient of functions is the quotient of their limits. So, we can find the limits of the
numerator and denominator separately, should they exist. And as the limit of a sum of
functions is the sum of their limits, we can find the limits term-by-term.

Now, we have lots of limits to
evaluate but theyβre all of very simple terms. And we can make them simpler by
taking the constants outside the limits. As the limit of a constant times a
function is that constant times the limit of the function. And now, the vast majority of our
limits are of the form the limit of π₯ to the power of some negative number as π₯
approaches infinity.

What are the values of such
limits? Well, we can write π₯ to the
negative π as one over π₯ to the π. And for π greater than zero, the
value is zero. All these limits then are zero. And weβre only left with the two
limits to worry about, both of which are limits of the constants four and three. The limit of a constant function is
just that constant. And so, being careful to include
this minus sign, we see the answer is negative four over three.

Solving this problem was
straightforward because we only had constants and negative powers in the numerator
and denominator. And we know what the limit of a
negative power of π₯ is as π₯ approaches infinity; itβs zero. Letβs now see an example where we
donβt just have negative powers.

Find the limit of π₯ squared plus
three all over eight π₯ cubed plus nine π₯ plus one as π₯ approaches infinity.

Our first thought might be to use
the fact that the limit of a quotient is the quotient of the limits. This gives us the limit of π₯
squared plus three as π₯ approaches infinity over the limit of eight π₯ cubed plus
nine π₯ plus one as π₯ approaches infinity. But we run into problems because
neither limit is defined. In the numerator, as π₯ approaches
infinity, π₯ squared plus three doesnβt approach any real value, it just gets bigger
and bigger without bound. And the same thing happens in the
denominator. As π₯ increases without bound, the
cubic eight π₯ cubed plus nine π₯ plus one also increases without bound.

Or maybe you think both limits
should be infinity. And so, the limit on left-hand side
is infinity over infinity. But this, just like a zero over
zero, is an indeterminate form. And it doesnβt tell us the value of
our limit. We need to use a different
approach.

The trick to this question is to
find the highest power of π₯ that appears in the numerator or denominator. Thatβs π₯ cubed here. And having found this highest
power, we divide both numerator and denominator by it. What do we get? π₯ squared divided by π₯ cubed is
π₯ to the negative one. And three divided by π₯ cubed is
three π₯ to the negative three. And in the denominator eight π₯
cubed divided by π₯ cubed is just eight. Nine π₯ divided by π₯ cubed is nine
π₯ to the negative two. And one divided by π₯ cubed is π₯
to the negative three.

So, now, we just have negative
powers of π₯ and a constant in the numerator and denominator. And as a result, when we apply this
limit law, we find that the limits in the numerator and denominator do now
exist. Letβs find their values. We can use the fact that the limit
of a sum of functions is the sum of their limits. This allows us to find the limit of
each term separately. We can also take the coefficient
outside the limits.

And now, apart from one limit,
which is the limit of a constant function, and whose value must therefore be eight,
all the other limits have the form the limit of π₯ to the power of negative π as π₯
approaches infinity. Where π is, of course, greater
than zero. And we know the value of such
limits. The value is always zero.

So, this is zero, and this is zero,
and this is zero, and this is zero. Simplifying then, our answer is
zero over eight, which is, of course, just zero. Now, in the first failed attempt at
solving this problem, we said that the limits of the numerator and denominator
individually were both undefined, or infinite. What do we mean by saying that the
value of such limits can be infinite. Letβs find out.

If the limit of π of π₯, as π₯
approaches infinity, is infinity, this means that we can make the value of π of π₯
arbitrarily large by choosing π₯ to be large enough. Suppose you want π of π₯ to be
greater than a billion. Well, thereβs some value such that
if we choose π₯ to be greater than that value, then π of π₯ will be greater than a
billion as required. Another way to think about this is
that, beyond a certain point, as π₯ increases without bound, π of π₯ also increases
without bound.

Similarly, the limit of π of π₯,
as π₯ approaches infinity, being negative infinity means that as π₯ increases
without bound, π of π₯ decreases without bound. And for completeness, we write down
the meanings when π₯ tends to negative infinity as well. Letβs see an example.

Find the limit of six π₯ squared
over π₯ minus six as π₯ approaches infinity.

There are various ways to find this
limit. One way is to look at the graph of
π¦ equals six π₯ squared over π₯ minus six. It looks like, as π₯ increases
without bound, six π₯ squared over π₯ minus six also increases without bound. As a result, we can say that this
limit is infinity. But you might not be convinced by
this. Perhaps, the graph does something
slightly different further along the π₯-axis.

We can also perform a polynomial
long division to find out six π₯ squared over π₯ minus six equals six π₯ plus 36
plus 216 over π₯ minus six. And itβs straightforward to take
limits on the right-hand side. We can do this term-by-term. The limit of six π₯, as π₯
approaches infinity, must be infinity. As π₯ increases without bound, six
π₯ also increases without bound. The limit of 36, as π₯ approaches
infinity, is just 36. This is the limit of a
constant.

And the last limit might be a bit
more tricky. We divide numerator and denominator
by the highest power of π₯ that we see; thatβs π₯. The limit of a quotient is the
quotient of the limits. And the limit in the numerator is
just zero and in the denominator is just one. So, the value of this limit is
zero. So, our limit is infinity plus
36. And when weβre dealing with limits,
itβs perfectly fine to say that infinity plus 36 is just infinity, which gives
another path to this answer. Okay, now, letβs see a final
problem.

Find the limit of nine minus eight
π₯ plus six π₯ squared minus two π₯ cubed as π₯ approaches negative infinity.

The first thing we might be tempted
to do is to write this limit of a sum as the sum of some limits. We can then evaluate each of these
limits one-by-one. The limit of the constant function
nine is just nine. What can we say about the limit of
eight π₯ as π₯ approaches negative infinity? Well, with the graph of π¦ equals
eight π₯ in mind, we can see that as π₯ decreases without bound, π¦ also decreases
without bound. And so, the limit of eight π₯, as
π₯ approaches negative infinity, is negative infinity.

How about the limit of six π₯
squared as π₯ approaches negative infinity? Again, we have the graph in mind,
and we see that as π₯ decreases without bound, π¦ increases without bound. So, this limit is infinity. And finally, the limit of two π₯
cubed as π₯ approaches negative infinity, we know what a cubic curve looks like. And we see that as π₯ approaches
negative infinity, π¦ also approaches negative infinity. This limit is negative
infinity.

So, it looks like our limit is nine
minus negative infinity plus infinity minus negative infinity. And if we treat infinity like a
number, we can write minus negative infinity as plus infinity, getting nine plus
infinity plus infinity plus infinity. And this sum is equal to
infinity. Now, we have to be slightly careful
about manipulating infinity in this way. But it turns out that all of these
steps are okay in this situation. We got lucky though that we didnβt
have any minus signs left at the end, as infinity minus infinity is undefined.

For various reasons then, it might
be worth seeing how to solve this problem in a different way. What we do instead is we factor out
the highest power of π₯, thatβs π₯ cubed, from inside the limit. This gives us the limit of π₯ cubed
times nine π₯ to the negative three minus eight π₯ to the negative two plus six π₯
to the negative one minus two as π₯ approaches negative infinity. The limit of a product is the
product of the limits.

Now, whatβs the limit of π₯ cubed
as π₯ approaches negative infinity? Well, we can slightly modify our
graph and call this the graph of π¦ equals π₯ cubed instead. And weβll see that this limit is
negative infinity. How about this limit? Well, these terms with negative
powers of π₯ contribute nothing, and so weβre left with just the limit of negative
two as π₯ approaches negative infinity. And this is, of course, just
negative two. The only bit of infinity
manipulation we need to do is to multiply negative infinity by negative two. The minus signs cancel. And we get just infinity.

Alternatively, we couldβve factored
the whole term negative two π₯ cubed out of the limit. And then, the value of the second
limit in our product would just be one. We can easily show that the first
limit in the product is infinity. And you might be more willing to
believe that infinity times one is infinity than you were to believe that negative
infinity times negative two is infinity.

Using this method, we can show that
the limit of a polynomial, as π₯ approaches positive or negative infinity, is just
the limit of the highest-degree term of that polynomial as π₯ approaches positive or
negative infinity. Then, all we have to do is look at,
or imagine, a graph of this monomial function.

Letβs see the key points weβve
covered in this video. We can consider limits of the form
the limit of π of π₯ as π₯ approaches positive or negative infinity. And in these cases, the limit laws
still apply. The limit of the reciprocal
function one over π₯, as π₯ approaches either positive or negative infinity, is
zero. And hence, combining this with one
of the limit laws, we see that the limit of one over π₯ to the power of π, as π₯
approaches positive or negative infinity, is also zero if π is greater than
zero.

We can find the limits of rational
functions by dividing numerator and denominator by the highest power of π₯ and using
the above result. And in a similar way, we can show
that the limit of a polynomial, as π₯ approaches positive or negative infinity, is
just the limit of its highest-degree term. We need to be slightly careful when
playing with infinity. But with some exceptions, for
example, the indeterminate forms infinity over infinity and infinity minus infinity,
infinity can be manipulated like a real number in the context of limits.