### Video Transcript

Find the limit as π₯ goes to π
over two of two minus two sin of π₯ over four π₯ minus two π.

First, we try solving it by direct
substitution. However, weβll reach zero over
zero, which is undefined. So we must find this limit by
another means. Letβs start by cancelling a factor
of two in both the numerator and denominator of the fraction. Now letβs consider some of the
rules we know. When considering the first rule
here, we notice that the value inside of the sine must be the same as the value in
the denominator of the function. In our limit, we have sin of π₯ in
the numerator. However, in the denominator, we
have two π₯ minus π. And these two things are not
equal. Therefore, we cannot use this first
rule.

In order to use the second rule, we
require a cos π₯ in the numerator. However, in our limit, we currently
have a sine. Here weβll be using an identity in
order to change the sine to a cosine. We have that sin of π₯ is equal to
cos of π₯ minus π by two. And we can substitute this into our
limit. Factorizing the denominator of the
fraction here, we can see that this is now of a very similar form to the rule which
we know.

At this stage, we need to perform a
substitution. We will substitute in π’ is equal
to π₯ minus π by two. However, we first need to consider
what will happen to our limit. So thatβs π₯ approaching π by
two. Well, weβll simply consider what
happens to the value of π’ as π₯ approaches π by two. Our value of π’ will approach π by
two minus π by two, which is simply zero. And so now weβre ready to
substitute π’ is equal to π₯ minus π by two into our limit. We obtain the limit as π’
approaches zero of one minus cos of π’ over two π’. We can factor out the half. And now we notice that our limit is
identical to our rule. And so, therefore, it must equal
zero. And this gives us a solution of
zero.