Video Transcript
Given π of π₯ equals one over π₯,
if possible or necessary, define π of zero so that π is continuous at π₯ equals
zero.
So we have the reciprocal function
here, which hopefully we know and love. And we know what its graph looks
like. The graph has two pieces, one in
the first quadrant and the other in the third. And these pieces are separated by a
vertical asymptote at π₯ equals zero. Intuitively then, it feels like
this function is not continuous at π₯ equals zero. Because the graph of our function
is not one continuous curve but is formed of two curves with the break at that
asymptote π₯ equals zero. Just to the left of this line when
π₯ is small but negative, π of π₯ is larger magnitude but negative. And to the right of this line when
π₯ is small but positive, π of π₯ is larger magnitude and positive. So as we pass π₯ equals zero, the
right of π of π₯ changes from a value which is larger magnitude and negative to a
value which is larger magnitude and positive.
Letβs see if our intuition agrees
with the technical definition by going through our checklist to see if the function
is continuous. For π to be continuous at π₯
equals zero, we need π of zero to be defined. The limit of π of π₯ as π₯
approaches zero to exist. And finally, we need these values
to be equal. Okay, so is π defined at zero? Well, if we substitute zero into
the definition of π of π₯, π of π₯ is one over π₯, we get π of zero is one over
zero. And this is not defined. Zero is not in the domain of our
function. So π is not defined at zero. And hence, π is not continuous at
π₯ equals zero. But this isnβt necessarily a
surprise as our task is to define π of zero to make π continuous at π₯ equals
zero. We wouldnβt have any work to do if
π were continuous at π₯ equals zero and π of zero was defined.
We check the second criterion. The limit of π of π₯ as π₯
approaches zero must exist. With the idea that if this limit
does exist, then we can just define π of zero to be the value of this limit. And, then the function will be
continuous at π₯ equals zero as required. Does this element exist? Well, if we look at the left-hand
limit as π₯ approaches zero, π of π₯ approaches negative infinity. And it gets worse. The limit from the right is
positive infinity. As π₯ approaches zero from the
right, π of π₯ gets larger and larger without bound.
The limit of π of π₯ as π₯
approaches zero therefore does not exist. And this is a problem. We can define π of zero to be
whatever we like. But whatever we define it to be,
this doesnβt change the fact that the limit of π of π₯ as π₯ approaches zero does
not exist. And so π cannot be continuous at
π₯ equals zero. So this is our answer then. The limit of π of π₯ as π₯
approaches zero doesnβt exist. And so π cannot be made continuous
at π₯ equals zero by defining π of zero. The fact that π of zero is
undefined is not really the problem here. And looking at the graph, we can
kind of see why this is true. There isnβt just a small gap on the
graph, which can be plugged by a single point. There is a huge chasm. While π of π₯ is not continuous at
π₯ equals zero and cannot be made so just by defining π of zero, itβs worth
pointing out that the function π of π₯ is actually continuous at other values of
π₯.
Given π of π₯ equals one over π₯,
if possible or necessary, define π of one so that π is continuous at π₯ equals
one. We go through our checklist with
this new question. We need π of one to be
defined. The limit of π of π₯ as π₯
approaches one to exist. And the limit of π of π₯ as π₯
approaches one to equal π of one. Is π of one defined? Yes, π of one is one over one,
which is one. π of one is defined. One is in the domain of π. Now, does the limit of π of π₯ as
π₯ approaches one exist? Letβs look at the graph. As π₯ approaches one from the left
and as π₯ approaches one from the right. The value of π of π₯ approaches
the same at finite value, the value one. If we wanted to find the value of
this limit without using the graph, weβd probably just use direct substitution. We can see then that the third item
on the checklist at the limit of π of π₯ as π₯ approaches one equals π of one is
also satisfied. The values of the left-hand side
and right-hand side are both one. Our answer then is that π is
already defined and continuous at π₯ equals one. Itβs not necessary to define π of
one again to make this so.