### Video Transcript

Find the limit as 𝑥 approaches
zero of one minus the tan of five 𝑥 all divided by the sin of five 𝑥 minus the cos
of five 𝑥.

We’re asked to evaluate the limit
as 𝑥 approaches zero of a function. And the first thing we should
always check is, are we allowed to use direct substitution? And to do this, we’re going to need
to look at the function inside of our limit. First, we need to recall we’re
allowed to use direct substitution on any trigonometric function. And since five 𝑥 is a linear
function, this means we’re allowed to use direct substitution on the sin of five 𝑥,
the cos of five 𝑥, and the tan of five 𝑥.

But the function inside of our
limit is not just a trigonometric function. For example, we take the difference
of two trigonometric functions in our denominator. And we have one minus the tan of
five 𝑥 in our numerator. But we also know if we can evaluate
two functions by using direct substitution, we can then evaluate their difference by
using direct substitution. So we can evaluate the sin of five
𝑥 minus the cos of five 𝑥 by using direct substitution since it’s the difference
between two trigonometric functions. And the same is true in our
numerator. Since one is a constant, its limit
will just be equal to one. Therefore, we’ve shown we can
evaluate the limit of our numerator by using direct substitution and the limit of
our denominator by using direct substitution.

And the last thing we need to use
is, if we have two functions which we can evaluate their limit by using direct
substitution, we can also evaluate the limit of their quotient by using direct
substitution. This means we can evaluate the
limit given to us in the question by using direct substitution. So let’s try evaluating this limit
by using direct substitution.

Remember, this means we substitute
𝑥 is equal to zero into our function. Doing this gives us one minus the
tan of five times zero all divided by the sin of five times zero minus the cos of
five times zero. In our numerator, we get one minus
the tan of zero. Well, the tan of zero is equal to
zero. So our numerator simplifies to give
us one. And in our denominator, the sin of
zero is equal to zero and the cos of zero is equal to one. So our denominator simplifies to
give us negative one. And we can calculate that this is
equal to negative one.

Therefore, we were able to show
we’re allowed to evaluate the limit given to us in the question by using direct
substitution. And then by doing this, we found
that the limit as 𝑥 approaches zero of one minus the tan of five 𝑥 all divided by
the sin of five 𝑥 minus the cos of five 𝑥 is equal to negative one.