Video: Finding Limits Involving Trigonometric Functions

Determine lim_(π‘₯ β†’ 0) (9 + (1/(4π‘₯))) sin 2π‘₯.

03:21

Video Transcript

Determine the limit as π‘₯ approaches zero of nine plus one divided by four π‘₯ multiplied by the sin of two π‘₯.

We’re asked to evaluate the limit as π‘₯ 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 π‘₯ 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 can’t use direct substitution if our denominator evaluates to give us zero. So, we’re 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 we’re 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 π‘₯ approaches zero of the sin of π‘₯ divided by π‘₯ is equal to one. So, let’s try rewriting our limit.

We’ll start by distributing the sin of two π‘₯ over our parentheses. This gives us the limit as π‘₯ approaches zero of nine sin of two π‘₯ plus the sin of two π‘₯ divided by four π‘₯. And now, we can see we can evaluate the limit as π‘₯ approaches zero of nine times the sin of two π‘₯ 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 we’re taking the sin of two π‘₯ instead of the sin of π‘₯.

And we know a few different ways of rewriting the sin of two π‘₯ to be in terms of the sin of π‘₯. For example, we could use the double-angle formula, and this would work. However, we’re going to rewrite our limit rule to be in terms of the sin of two π‘₯. We’ll start by replacing all instances of π‘₯ with two π‘₯. This gives us the limit as two π‘₯ approaches zero of the sin of two π‘₯ divided by two π‘₯ is equal to one.

We have to be careful here, since the limit we’re trying to evaluate has π‘₯ approaching zero. However, the limit we now have has two π‘₯ approaching zero. But if two π‘₯ is approaching zero, then π‘₯ must be getting smaller and smaller. In fact, π‘₯ must be approaching zero. So, we have the limit as π‘₯ approaches zero of the sin of two π‘₯ divided by two π‘₯ is equal to one.

So, to use our newfound limit rule, we’ll 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 π‘₯ 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 π‘₯ in our denominator instead of two π‘₯. 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, we’ve shown the limit as π‘₯ approaches zero of nine plus one divided by four π‘₯ multiplied by the sin of two π‘₯ is equal to one-half.

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