# Video: Finding the Limit of a Quotient of Trigonometric and Linear Functions at a Point

Determine lim_(𝑥 → (π/4)) ((9 sin 𝑥)/5𝑥).

07:44

### Video Transcript

Determine the limit of the function nine sine 𝑥 divided by five 𝑥 as 𝑥 approaches 𝜋 over four.

We know that for a function 𝐹, we can use direct substitution to evaluate the limit of 𝑓 if 𝑓 is a trigonometric, exponential, polynomial, or logarithmic function. And if 𝑓 is the quotient of functions for which direct substitution works. Since 𝑓 is the quotient of a trigonometric function and a polynomial, we can evaluate our limit by using direct substitution. Since our limit has 𝑥 approaching 𝜋 over four, we would substitute the value of 𝜋 over four into our function and evaluate the result. This gives us the expression nine multiplied by sin of 𝜋 over four divided by five multiplied by 𝜋 over four. We can then evaluate this expression to get nine root two over two divided by five 𝜋 over four.

We can then use our rule that dividing by a fraction is the same as inverting the fraction and multiplying to see that our limit is equal to nine root two over two multiplied by four divided by five 𝜋. We can simplify this since both the numerator and the denominator have a factor of two. Finally, we can multiply these two fractions together to give us that the limit of nine sin 𝑥 divided by five 𝑥 as 𝑥 approaches 𝜋 over four is equal to 18 root two divided by five 𝜋. Now, we could end here. But let’s clear some space and try to justify why we can use direct substitution to evaluate this limit.

If we were to attempt to solve this problem without using direct substitution, we could be tempted to use the quotient rule for limits, which states that for any functions 𝑓 of 𝑥 and 𝑔 of 𝑥, if the limit of 𝑔 of 𝑥 as 𝑥 approaches 𝑎 is not zero, then the limit of the quotient is equal to the quotient of the limits. In our case, we want 𝑔 of 𝑥 to be equal to five 𝑥 and 𝑎 to be equal to 𝜋 over four. Therefore, to justify being able to use the quotient rule for limits, we need that the limit of five 𝑥 as 𝑥 approaches 𝜋 over four must not be equal to zero. Okay, so let’s try to evaluate this limit without using direct substitution.

We know that for any two functions. The products of the limit is equal to the limit of the products. So we could split our limit of five multiplied by 𝑥 into the limit of five as 𝑥 approaches 𝜋 over four multiplied by the limit of 𝑥 as 𝑥 approaches 𝜋 over four. Next, we know that for any constant 𝐾, the limit as 𝑥 approaches 𝑎 of 𝐾 is just equal to 𝐾. We can apply this to evaluate the limit of five as 𝑥 approaches 𝜋 over four to just be equal to five. We also know that the limit of the function 𝑥 as 𝑥 approaches 𝑎 is just equal to 𝑎.

Next, we can apply this to see that the limit of 𝑥 as 𝑥 is approaching 𝜋 over four is equal to just 𝜋 over four. We can now see that the limit of our function five 𝑥 as 𝑥 approaches 𝜋 over four is not equal to zero. This justifies our usage of the quotient rule for limits. So now that we’ve justified being able to use the quotient rule for limits, let’s clear some space and go back to the limit in the question.

We start by rewriting our limit as the quotient of these two limits. Now we want to use the fact that the limit of a product is equal to the product of a limit to rewrite both the limit in the numerator and the limit in the denominator. We can split the numerator. And it’s the limit of nine as 𝑥 approaches 𝜋 over four multiplied by the limit of sin 𝑥 as 𝑥 approaches 𝜋 over four. We can do the same to split the denominator, giving us the limit of five as 𝑥 approaches 𝜋 over four multiplied by the limit of 𝑥 as 𝑥 approaches 𝜋 over four.

Now we can use the fact that the limit of any constant 𝐾 is just equal to 𝐾 to evaluate the limit of nine in the numerator and the limit of five in the denominator. This gives us nine multiplied by the limit of sin 𝑥 as 𝑥 approaches 𝜋 over four divided by five multiplied by the limit of 𝑥 as 𝑥 approaches 𝜋 over four. Next, we can use the fact that the limit of the function 𝑥 as 𝑥 approaches 𝑎 is equal to 𝑎 to evaluate our limit of 𝑥 in our denominator. So we can replace our limit of 𝑥 with just 𝜋 over four. This means that the last part of our question that we need to evaluate is the limit of sin 𝑥 as 𝑥 approaches 𝜋 over four.

We can recall that the limit being evaluated by direct substitution is the same as saying that the function sin 𝑥 is continuous at 𝜋 over four. But we do not need to worry about this here. We can attempt to justify our use of direct substitution by looking at a function table for sin 𝑥. On the top row, we will have our input values of 𝑥. And on the bottom row, we will have the outputs of the function sin 𝑥. We will include an entry for 𝜋 by four in our table. But since this is the value we’re attempting to evaluate, we will leave its output blank. We add in some input values of 𝑥 around 𝜋 over four, which get progressively close from closer to 𝜋 over four. We can evaluate the outputs of our function by substituting our input values of 𝑥 into the function sin 𝑥. This gives us the following table.

We know that the limit of a function 𝑓 of 𝑥 as 𝑥 approaches 𝑎 is equal to 𝐿 if the limit of 𝑓 of 𝑥 as 𝑥 approaches 𝑎 from both the left and the right is equal to 𝐿. So to evaluate our limit of sin 𝑥 as 𝑥 approaches 𝜋 over four, we will need to evaluate the left- and right-hand limit. We can use the following values in our table to evaluate the limit from the left. We write these outputs in a sequence. And we can see that as our input is getting closer and closer to 𝜋 over four, our outputs are approaching the square root of two over two. We can do the same for the right-hand limit. And we can see that as our inputs are approaching 𝜋 over four from the right, our outputs are getting closer and closer to the square root of two over two. Hence, the limits of sin 𝑥 as 𝑥 approaches 𝜋 by four from the left and from the right are equal to the square root of two over two.

Therefore, since the left- and right-hand limit are both equal to the square root of two over two, we must have that the limit of sin 𝑥 as 𝑥 approaches 𝜋 over four is equal to the square root of two over two. We can then replace the limit of sin 𝑥 in our expression with the square root of two over two. This gives us that our limit is equal to nine multiplied by the square root of two over two divided by five multiplied by 𝜋 over four. Finally, if we were to evaluate this expression, we would get that the limit of nine sin 𝑥 over five 𝑥 as 𝑥 approaches 𝜋 over four is equal to 18 root two divided by five 𝜋.