### Video Transcript

Evaluate the limit of sin π₯ over
sin π₯ over two as π₯ approaches zero.

Letβs take a look at this
limit. The first thing we should try is
direct substitution. We substitute the number that π₯ is
approaching, zero, for π₯ in the expression. And doing this, we get sin zero
over sin zero over two. The sin of zero is zero, so have a
zero in the numerator. And also of course sin of zero over
two is just sin of zero, which is zero. And so we have zero in the
denominator as well.

Zero over zero is an indeterminate
form. And so unfortunately direct
substitution hasnβt given us the value of the limit that weβre looking for. Still it was worth a try. So how do you evaluate this
limit? Well weβre going to see two methods
in this video. The first method makes use of the
double-angle identity for sine, that sin two π is two times sin π times cos π for
any angle π.

And if we choose to substitute π₯
over two for π, we find that sin of two times π₯ over two is two sin π₯ over two
times cos π₯ over two. And of course on the left-hand
side, two times π₯ over two is just π₯. So we find that sin π₯ is two sin
π₯ over two cos π₯ over two. Okay, but how does this help
us? Well, if we use the expression in
our limit, we see that we have a factor of sin π₯ over two in the numerator, which
will cancel with the sin π₯ over two in the denominator. Upon the cancelling, what do we
get?

We get the limit of two cos π₯ over
two as π₯ approaches zero. And this is a limit that we can
evaluate using direct substitution. Replacing π₯ by zero, we get two
cos zero over two. And cos zero over two is just cos
zero, which is one. And so our answer is two times one,
which is two. This method required using a clever
substitution for sin π₯: a substitution which may not have been obvious, but once
weβve made it, it was clear what to do.

We can however use a different
trick to evaluate this limit. This method uses the fact that the
limit of sin π₯ over π₯ as π₯ approaches zero is one. This is an important limit as it
allows us to differentiate sine. If you havenβt seen this limit
before, then itβs probably best to stop the video here and make do with the first
method. But if you have seen this limit
before, you might be interested in how it can help us solve our problem. Well we can write sin π₯ over sin
π₯ over two as sin π₯ over π₯ times π₯ over sin π₯ over two.

You can check that this is right by
multiplying out the fractions on the right-hand side and then simplifying. Basically what weβve done is to
divide and then multiply by π₯. And we can then use the fact that
the limit of a product of functions is the product of their limits. And so we can write this limit as
the product of two limits. And one of these limits we know the
value of: the limit of sin π₯ over π₯ as π₯ approaches zero is one. And so we only have the other limit
to worry about: the limit of π₯ over sin π₯ over two as π₯ approaches zero.

How do we find this limit? Well itβs very nearly the
reciprocal of the limit we know. The limits of the reciprocal of
some function is just the reciprocal of the limit of that function. And so we see that the limit of π₯
over sin π₯ as π₯ approaches zero is one. And thereβs nothing particularly
special about the use of the letter π₯ here. We could use any letter, perhaps
π. Or thinking of the limits we have,
we could set π equal to π₯ over two. The limit of π₯ over two over sin
π₯ over two as π₯ over two approaches zero is one.

You might feel slightly uneasy
about replacing π by something that isnβt just another letter, but this can be
justified. What does it mean for π₯ over two
to approach zero? Well this just means that π₯ itself
is approaching zero. And so we have that the limit of π₯
over two over sin π₯ over two as π₯ approaches zero is one. This is really close to the limit
we have. The only difference being that we
have π₯ in the numerator rather than π₯ over two. But we can write that π₯ in the
numerator as two times π₯ over two, and we know that the limit of two times
something is two times the limit of that something.

Now we have a limit whose value we
know. The value of this limit is one. The value of the limit in our
question is therefore two times one, which is two. And this is our final answer. This method didnβt require using a
clever identity for sin π₯. In general, using the limit of sin
π₯ over π₯ as π₯ approaches zero and the corresponding limits for cosine and tan
allows you to rewrite a trigonometric limit as a limit of a rational function, which
you may find easier to solve.