Video Transcript
Determine the limit as ๐ฅ approaches zero of nine plus one divided by four ๐ฅ multiplied by the sin of two ๐ฅ.
Weโre asked to evaluate the limit as ๐ฅ approaches zero of a product between a rational function and a trigonometric function. And we can attempt to evaluate rational functions and trigonometric functions by direct substitution. So, we can attempt to evaluate their products by direct substitution.
Substituting ๐ฅ is equal to zero, we get nine plus one divided by four times zero multiplied by the sin of two times zero. And we see that this is a problem. This simplifies to give us nine plus one divided by zero all multiplied by zero. And we know we canโt use direct substitution if our denominator evaluates to give us zero. So, weโre going to need to find a different way to evaluate this limit.
We could try and rewrite this limit in terms of limits we do know how to evaluate. We see in the limit weโre asked to evaluate we have the sine function divided by a linear function. And this reminds us of a useful trigonometric limit result which we should commit to memory: the limit as ๐ฅ approaches zero of the sin of ๐ฅ divided by ๐ฅ is equal to one. So, letโs try rewriting our limit.
Weโll start by distributing the sin of two ๐ฅ over our parentheses. This gives us the limit as ๐ฅ approaches zero of nine sin of two ๐ฅ plus the sin of two ๐ฅ divided by four ๐ฅ. And now, we can see we can evaluate the limit as ๐ฅ approaches zero of nine times the sin of two ๐ฅ by using direct substitution. And our second term is almost in a form which we already know the limit of. We just have a constant factor of one-quarter, and weโre taking the sin of two ๐ฅ instead of the sin of ๐ฅ.
And we know a few different ways of rewriting the sin of two ๐ฅ to be in terms of the sin of ๐ฅ. For example, we could use the double-angle formula, and this would work. However, weโre going to rewrite our limit rule to be in terms of the sin of two ๐ฅ. Weโll start by replacing all instances of ๐ฅ with two ๐ฅ. This gives us the limit as two ๐ฅ approaches zero of the sin of two ๐ฅ divided by two ๐ฅ is equal to one.
We have to be careful here, since the limit weโre trying to evaluate has ๐ฅ approaching zero. However, the limit we now have has two ๐ฅ approaching zero. But if two ๐ฅ is approaching zero, then ๐ฅ must be getting smaller and smaller. In fact, ๐ฅ must be approaching zero. So, we have the limit as ๐ฅ approaches zero of the sin of two ๐ฅ divided by two ๐ฅ is equal to one.
So, to use our newfound limit rule, weโll start by rewriting our limit of a sum as the sum of two limits. As we said before, we can evaluate the first limit by using direct substitution. Substituting ๐ฅ is equal to zero, we get nine times the sin of two times zero. But the sin of zero is just equal to zero. So, this limit just evaluates to give us zero.
We want to evaluate our second limit by using our limit rule. However, we see that we have four ๐ฅ in our denominator instead of two ๐ฅ. But remember, four is just a constant. So, we could take out a factor of one-half from our limit. And we now see we have one-half multiplied by our limit rule. So, our second term evaluates to give us one-half multiplied by one, which is just equal to one-half.
So, weโve shown the limit as ๐ฅ approaches zero of nine plus one divided by four ๐ฅ multiplied by the sin of two ๐ฅ is equal to one-half.
