### Video Transcript

The Existence of Limits

In this video, we will learn how to
determine whether the limit of a function at a certain value exist. There are some reasons why a limit
may not exist. We will be looking at these reasons
and using them to deduce whether or not certain limits exist. When we are considering the limit
as π₯ approaches π of π of π₯, there are a few reasons why the limit may not
exist. The first is if the limit is not
finite. In the case of this π of π₯, we
can see that as π₯ approaches zero, π of π₯ is approaching positive or negative
infinity depending on whether weβre approaching zero from the right or from the
left. The reason why this limit does not
exist is because the limit must approach a particular point. And we cannot say that infinity is
a point since infinity does not actually exist; itβs just a concept.

Letβs quickly note some special
cases when we have an infinite limit. In the case of π of π₯, we can see
that the left and right limits, as π₯ approaches zero, both tend to positive
infinity. Since these left and right limits
both tend to the same sign of infinity, we can say that the limit as π₯ approaches
zero of π of π₯ is also equal to infinity. However, we must note that this
limit still does not exist since infinity is not a number; itβs just a concept. Similarly, we can note that, for β
of π₯, the left and right limits as β approaches zero both tend to negative
infinity. And so we can say that the limit as
π₯ approaches zero of β of π₯ is equal to negative infinity. But again, this limit still does
not exist. Though it is important to remember
these as it can be useful to know whether a particular limit tends to a certain sign
of infinity, rather than the case of π of π₯ where the different sides of the limit
tend to different signs of infinity.

Letβs now consider another case
which can cause a limit not to exist. We can say that the limit as π₯
approaches π of π of π₯ does not exist if the limit does not approach a particular
point. And this happens with certain
oscillating functions. Here is an example of an
oscillating function π of π₯ where a limit does not exist. And this limit is as π₯ approaches
zero. We can see that the closer the
values of π₯ get to zero, the more rapidly our function oscillates up and down. This function is oscillating
between the values of minus one and one. The closer π₯ gets to zero, the
quicker it oscillates between minus one and one. And so we can say that the function
does not approach a particular value as it approaches zero, since the closer and
closer it gets to zero, the faster it changes between one and minus one. And from this, we can say that π
of π₯ does not approach a particular point as π₯ approaches zero. And, therefore, the limit as π₯
approaches zero of π of π₯ does not exist.

Now, letβs consider another reason
why limits may not exist. When finding the limit as π₯
approaches π of π of π₯, itβs very important to consider the left and right
limits. This is because if the left or the
right limit does not exist or if both the left and right limits exist but they are
not equal, then the limit as π₯ approaches π of π of π₯ does not exist. If we consider the function π of
π₯, we can see that π of π₯ is a piecewise function which is defined between the
values π and π. Letβs consider the limit of π of
π₯ as π₯ approaches π. We can see that the limit as π₯
approaches π from the right is equal to π of π. However, as π₯ approaches π from
the left, π is undefined. And so, therefore, this limit from
the left does not exist.

This tells us that the limit as π₯
approaches π of π of π₯ does not exist. Now letβs consider the limit as π₯
approaches π of π of π₯. In order to consider this limit,
letβs label the two sections of our graph. Weβll call the section between π
and π π of π₯ and the section between π and π β of π₯. The limit as π₯ approaches π from
the right will be equal to β of π since we are coming from the right of π and the
value of π of π₯ will be moving along β of π₯.

Now, the limit as π₯ approaches π
from the left will be equal to π of π since as the value of π of π₯ gets closer
and closer to π from the left, it will be following along π of π₯. And so itβll be equal to π of
π. Since β of π and π of π are not
equal to one another, we must conclude that the limit as π₯ approaches π of π of
π₯ does not exist. Now, we have covered all of the
reasons why a limit may not exist. We can say that the limit as π₯
approaches π of π of π₯ exists if both the left and right limits exist and the
left limit is equal to the right limit. We can also say that the limit as
π₯ approaches π of π of π₯ is equal to some constant πΏ where πΏ is also equal to
the left and right limits. Weβre now ready to look at some
examples.

Discuss the existence of the limit
as π₯ approaches seven of π of π₯ given π of π₯ is equal to 13π₯ plus seven if one
is less than π₯ which is less than seven and 14π₯ plus seven if seven is less than
or equal to π₯ which is less than eight.

In this example, our π of π₯ is a
piecewise function. And we are asked to find the limit
as π₯ approaches seven. Seven is the π₯ value at which our
function switches between 13π₯ plus seven and 14π₯ plus seven. In order to find whether our limit
exists, we need to check whether the left and right limits exist and if they are
equal. Weβll start by considering the left
limit. Since π₯ is approaching seven from
below, we know that π₯ must be less than seven. Since π₯ is less than seven, we can
see from our piecewise definition that π of π₯ is equal to 13π₯ plus seven.

Since this is a polynomial, we can
use direct substitution. In order to find this limit, we
simply substitute π₯ equals seven into 13π₯ plus seven. And this gives us that the left
limit is equal to 98. Since the limit here is equal to a
real constant, we know that this limit must exist. Letβs now consider the limit as π₯
approaches seven from above. Since π₯ is approaching seven from
above, we have that π₯ is greater than seven. Therefore, from our piecewise
definition, we have that π of π₯ is equal to 14π₯ plus seven which is again a
polynomial. And so we can use direct
substitution in order to find this limit. Substituting π₯ equals seven into
14π₯ plus seven, we obtain the limit as π₯ approaches seven from the right is equal
to 105. Therefore, our limit exists.

Now, we have found that both the
left and right limit exist. However, the left limit is equal to
98. And the right limit is equal to
105. Therefore, we can conclude that the
limit as π₯ approaches seven of π of π₯ does not exist because the left and the
right limit are not equal to each other. In this example, we saw how the
limit did not exist because the left and right limits were not equal. This is because there is a jump in
the function at the point which weβre trying to take the limit. Therefore, we cannot say that the
limit of π of π₯ approaches a particular point since it depends upon which
direction we are approaching the limit as to what the limit could equal. Now, letβs consider another
example.

Discuss the existence of the limit
as π₯ approaches three of π of π₯ given π of π₯ is equal to the mod of π₯ minus
two plus three if negative two is less than π₯ which is less than three π₯ squared
plus six π₯ minus 27 all over π₯ squared minus three π₯ if three is less than π₯
which is less than nine.

Letβs consider the left and right
limits as π₯ approaches three of this piecewise function. As π₯ approaches three from below,
we know the π₯ will be less than three. Therefore, π of π₯ will be equal
to the mod of π₯ minus two plus three. We can use direct substitution in
order to find this limit. We find that it is equal to
four. Now, letβs consider the right-hand
limit. Since π₯ is approaching three from
above, we know that π₯ will be greater than three. So π of π₯ is equal to π₯ squared
plus six π₯ minus 27 all over π₯ squared minus three π₯. If we try using direct substitution
to find this limit, we will find that itβs equal to zero over zero which is
undefined. However, since this is a rational
function with polynomials in the numerator and denominator where both the numerator
and denominator are equal to zero at three, this means we can factorise out a factor
of π₯ minus three from the top and bottom.

We obtain π₯ plus nine multiplied
by π₯ minus three over π₯ multiplied by π₯ minus three. Next, we can simply cancel the π₯
minus three on the top and bottom of the fraction. However, we must be careful
because, in doing this, we are actually changing the domain of π of π₯. Previously inputting a value of π₯
equals three will have given us an undefined outcome. However, now if we input π₯ is
equal to three, we will get a number. Therefore, we should call this new
function π of π₯ where π of π₯ is identical to π of π₯ in every way except by its
domain, since we can input π₯ equals three into π of π₯, however, we cannot into π
of π₯. We are now able to use direct
substitution on π of π₯ in order to find the limit from above.

We obtain that the limit from above
is equal to four. Comparing this to the limit from
below, we can see that these two are equal. We can, therefore, conclude that
the limit as π₯ approaches three of π of π₯ exists. And it is equal to four. In this example, we saw how a limit
of a piecewise function at the point where the definition of the function changes
can exist if the left and right limit can agree on a value of which the limit is
equal to. Letβs now look at another
example.

Discuss the existence of the limit
as π₯ approaches two of one over the mod of π₯ minus two.

Letβs start by considering the
limit as π₯ approaches two of mod π₯ minus two which is the denominator of our
fraction. We can find this by using direct
substitution. And it gives us zero. What this tells us is that, as π₯
approaches two, the modulus of π₯ minus two gets smaller and smaller as it
approaches zero. And since mod of π₯ minus two is
getting smaller, this means that the reciprocal one over mod of π₯ minus two will be
getting larger and larger. And this tells us that here we have
an infinite limit. Therefore, the limit does not
approach a particular value and so the limit does not exist. This is because infinity is not a
number; itβs simply a concept. Now, there is more that we can
deduce other than the limit simply does not exist. So letβs draw the graph of one over
the mod of π₯ minus two.

Our graph will go to infinity at π₯
is equal to two. And we know that the graph must be
positive everywhere on π₯ since our graph has a modulus or absolute value within
it. Since everything outside of the
modulus is positive, this means that our graph must be positive everywhere on
π₯. We can see for our graph that, as
π₯ approaches two from above and below, our function is tending to positive
infinity. So we can say that both the left
and right limits as π₯ approaches two are equal to infinity. Since these are both equal to
positive infinity, we can conclude that the limit as π₯ approaches two of one over
mod of π₯ minus two is also equal to positive infinity.

Itβs important to remember that
this limit still does not exist since infinity is not a number. Therefore, we can conclude that the
limit as π₯ approaches two of one over the mod of π₯ minus two does not exist. But the limit is equal to
infinity. In this example, we saw what
happens when our limit tends to infinity. Now, letβs look at one final
example.

Investigate the limit of π of π₯
is equal to two cos of one over π₯ as π₯ tends to zero. (a) Complete the table of
values π of π₯ for values of π₯ that get closer to zero.

In order to solve this first part,
we simply substitute the values of π₯ into π of π₯. We find that π of one over 99 is
equal to negative two. π of one over 100π is equal to
two. Continuing in this way, we can
complete the rest of the table and we find that this is the solution to part a).

Part b), what does this suggest
about the graph of π close to zero?

Our table suggests that the closer
and closer we get to zero, the more rapidly the graph of π of π₯ will oscillate
between negative two and two. Letβs quickly consider why this is
the case. We know that as π₯ approaches zero
from above, one over π₯ would approach infinity. And as π₯ approaches zero from
below, one over π₯ would approach negative infinity. What this means is as π₯ gets
smaller, one over π₯ will get very very large in either the positive or negative
direction depending on which way weβre approaching zero. Now, the closer and closer we get
to zero, the rate at which one over π₯ will get larger increases.

Now, we know that the cos function
oscillates between negative one and one at a constant rate. However, since the rate at which
one over π₯ is increasing is increasing, this means that cos of one over π₯ will
oscillate between negative one and one at an increasing rate. So this is why our graph will
oscillate rapidly between negative two and two.

Part c), hence evaluate the limit
as π₯ approaches zero of π of π₯.

In part b), we saw that as π₯
approaches zero, π of π₯ will oscillate more and more rapidly between negative two
and two. So as π₯ approaches zero, we cannot
deduce a particular value of which π of π₯ will approach since the graph will be
changing between negative two and two more and more quickly. Since the function does not tend to
a particular point as π₯ goes to zero, we can deduce that the limit does not
exist. We have now seen why the limit of
an oscillating function may not exist. And so we have covered a few
examples of why a limit may or may not exist. Letβs recap some of the key points
of the video.

If a function tends to infinity at
a point then the limit of the function at that point does not exist. In order for a limit to exist it
must approach a particular point. So in the case of some oscillating
functions, they may begin to oscillate incredibly rapidly as they approach a
point. And, therefore, the limit will not
exist. If the limit as π₯ approaches π
from below of π of π₯ or the limit as π₯ approaches π from above of π of π₯ does
not exist or if the limit from below is not equal to the limit from above, then the
limit as π₯ approaches π of π of π₯ does not exist. If the limit as π₯ approaches π
from below of π of π₯ and the limit as π₯ approaches π from above of π of π₯ both
exist and the limit from below is equal to the limit from above which is equal to
some constant πΏ, then the limit as π₯ approaches π of π of π₯ exists. And we can say that the limit as π₯
approaches π of π of π₯ is equal to πΏ.