Question Video: Finding the Limit of a Quotient of Trigonometric Functions at a Point Mathematics

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.