### Video Transcript

Investigate the behavior of π of
π₯ equals two cos one over π₯ as π₯ tends to zero.

The first part of this question is
to complete the table of values of π of π₯ for values of π₯ that get closer to
zero. And if we look at the values of π₯
that weβre given in the table, we can see that they really do get closer and closer
to zero. One over 99π is already quite
small. But one over 100π is even smaller,
even closer to zero. And the trend continues, one over
999π is smaller still. Okay, now letβs go about completing
this table of values.

In the first cell we have to fill,
we need to put the value of π of one over 99π. But, what is this value? We find it by using the definition
of the function π. π of π₯ is two cos one over
π₯. And so, π of one over 99π is two
cos one over one over 99π. One over one over 99π is just
99π. And so, we can simplify to get two
cos 99π. Whatβs the value of cos of
99π? The function cos is a periodic
function. And its period β assuming weβre
working in radians, which we are β is two π. Another way of saying this is that
the value of cos of π₯ plus two π is the same as the value of cos of π₯. And by applying this rule twice, we
can see that cos π₯ plus four π is equal to cos π₯ plus two π which is equal to
cos π₯.

By repeating this process, we can
see that cos π₯ plus two ππ, for any natural number π, is equal to cos π₯. And in fact, by working backwards,
we can show that this holds for negative integers π too. And so, the result holds for any
integer π. Okay, but how does this result help
us simplify two cos 99π? Well, we can write 99π as π plus
two π times 49. And so, using our rule, we get two
cos π. Another way of thinking about this
is that if you keep subtracting two π from 99π, eventually, youβll end up with
just π. As cos π is just negative one, two
times cos π is negative two. And we found the value of π of one
over 99π.

Now, we need to find the next value
in our table. The process is almost exactly the
same. We use the definition of π and the
fact that one over one over 100π is 100π. But this time, as 100π is an even
multiple of π, we eventually get down to zero when we keep subtracting two π from
it. We therefore find that π of one
over 100π is two cos zero. And as cos zero is one, two cos
zero is two. We put this two into our table and
move on to the next entry.

π of one over 999π is two cos one
over one over 999π which is two cos 999π. And as 999 is an odd number, the
value of cos 999π is the value of cos π. And just like with π of one over
99π, as cos of π is negative one, π of one over 999π is negative two. Rather than using the same
procedure to fill in the remaining three gaps in our table, weβre going to use the
pattern that weβve discovered.

When π₯ is a unit fraction with an
odd multiple of π in its denominator, then the value of π of π₯ is negative
two. We can prove this if weβd like
to. Any odd number can be expressed as
two π plus one, where π is some natural number. And following our procedure, we can
see that π of one over two π plus one π is negative two. Pause the video if youβd like to
think about this some more.

And so, in particular, as 9999 is
an odd number, when π₯ is one over 9999π, then π of π₯ is negative two.

So that covers the case when the
denominator of our unit fraction is an odd multiple of π. But what about the even multiples
of π? Writing our general even number as
two times π, we see that when π₯ is one over an even multiple of π, π of π₯ is
two. And so, both π of one over 1000π
and π of one over 10000π are both two. We might also like to summarise the
discoveries weβve made, that π of one over two ππ is two and π of one over two
π plus one π is negative two, for any natural number π.

Letβs now move on to the next part
of the question. What does this suggest about the
graph of π close to zero? Well, weβve seen for some very
small values of π₯ in our table that π of π₯ can be negative two or two. And weβve seen from our discoveries
that π of one over two ππ is two and π of one over two π plus one π is
negative two. That however close you are to zero,
you can find a smaller value of π₯ for which π of π₯ is two or π of π₯ is negative
two. So what does this suggest about the
graph of π close to zero? It suggests that it oscillates
between negative two and two.

You might like to just check that
this makes sense, given the definition of our function π, that π of π₯ equals two
cos one over π₯. Whatever the value of π₯ is, one
over π₯ is just some number. And so, π of π₯ is two times cos
something which is bounded between negative two and two, inclusive. So certainly, our function canβt be
oscillating between two values greater in absolute value. Also, looking at the table you
might be tempted to think that π of π₯ can only take two possible values, negative
two or two. But actually, the function takes
values between negative two and two as well. Itβs just that the values of π₯ in
our table were chosen to produce these extreme values. You might like to think about how
to find values of π₯ for which π of π₯ is zero or one or negative 1.5.

Moving on to the third and final
part of the question, hence, evaluate the limit of π of π₯ as π₯ approaches
zero. Well, weβve seen that as π₯
approaches or tends to zero, π of π₯ oscillates between negative two and two. The value of π of π₯, therefore,
is not getting closer and closer to some real number. And so, the limit of π of π₯ as π₯
approaches zero cannot be any real number. And as π is a bounded function,
taking values between negative two and two, the limits cannot be infinite
either. The only conclusion we can draw is
that the limit does not exist.

Now, the purpose of this question
was to work all this out without relying on a graph. But now that we have worked it out,
you might like to use a graphing calculator or graphing software to explore the
graph of π. As you might be able to see from
the small graph Iβm showing on screen, something pretty wild happens near the
origin.