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Question Video: Finding Limits Involving Trigonometric Functions Mathematics • 12th Grade

Find lim_(π‘₯ β†’ 0) ((9π‘₯/sin 10π‘₯) βˆ’ (tan 2π‘₯/2π‘₯)).

03:44

Video Transcript

Find the limit as π‘₯ approaches zero of nine π‘₯ divided by the sin of 10π‘₯ minus the tan of two π‘₯ divided by two π‘₯.

In this question, we’re asked to evaluate the limit of the difference of two quotients. The first quotient is a linear function divided by a trigonometric function. And the second term is a trigonometric function divided by a linear function. And there’s a few different ways we could go about evaluating this limit. Since we can evaluate all of the parts of this limit by using direct substitution, we can start by attempting to evaluate this limit by direct substitution.

Substituting π‘₯ is equal to zero into our function gives us nine times zero divided by the sin of 10 times zero minus the tan of two multiplied by zero divided by two times zero. And if we evaluate the numerators and denominators of each term separately, we can see we get zero divided by zero minus zero divided by zero.

In particular, this tells us the limit of the first function is an indeterminate form by using direct substitution. And the limit of the second term is also an indeterminate form by using direct substitution. So we can’t evaluate the limit of either term by direct substitution. So instead, we’re going to need to use a different method. Since each term is the quotient of a linear function and a trigonometric function, we’ll do this by recalling some of our useful trigonometric limit results.

In particular, we can recall for any real constant π‘Ž, the limit as π‘₯ approaches zero of the sin of π‘Žπ‘₯ divided by π‘₯ is equal to π‘Ž. And for any real constant π‘Ž, the limit as π‘₯ approaches zero of the tan of π‘Žπ‘₯ divided by π‘₯ is also equal to π‘Ž. And these two terms are almost in the exact form we need. We just need to rearrange them slightly by using our properties of the limit.

Let’s start with the first term. We can see we’re dividing by the sin of 10π‘₯. However, in our limit result, we have the sin of π‘Žπ‘₯ in the numerator. We can get around this by taking the reciprocal of both sides of our limit results. We’ll need to use the power rule for limits to take the reciprocal inside of our limit. And of course, this means π‘Ž is not allowed to be zero since we’re dividing by π‘Ž. This gives us the limit as π‘₯ approaches zero of π‘₯ divided by the sin of π‘Žπ‘₯ is equal to one divided by π‘Ž provided π‘Ž is nonzero.

We can now evaluate the limit of each term separately by using these limit results. So let’s start by taking the limit of a difference as the difference of the limits. This then gives us the limit as π‘₯ approaches zero of nine π‘₯ divided by the sin of 10π‘₯ minus the limit as π‘₯ approaches zero of the tan of two π‘₯ divided by two π‘₯. And it is worth noting, we can only split our limit as the difference of two limits if both of these two limits exist. But we can justify this by using our two limit results.

Let’s start with the first limit. We want to use the fact that the limit as π‘₯ approaches zero of π‘₯ divided by the sin of π‘Žπ‘₯ is one over π‘Ž. Our limit is almost in this form. We’re just going to need to take the constant factor of nine outside of our limit. This gives us nine times the limit as π‘₯ approaches zero of π‘₯ divided by the sin of 10π‘₯. Now, our value of π‘Ž is equal to 10. So this evaluates to give us one over 10.

We get a very similar story in our second limit. We want to use the fact that the limit as π‘₯ approaches zero of the tan of π‘Žπ‘₯ divided by π‘₯ is equal to π‘Ž. However, we don’t have π‘₯ in our denominator. Instead, we have two π‘₯. So we’re going to need to take a constant factor of one-half outside of our limit. Then, we can evaluate the limit as π‘₯ approaches zero of the tan of two π‘₯ divided by π‘₯ to be the coefficient of π‘₯, which is two.

So this just gives us nine multiplied by one-tenth minus a half multiplied by two. And we can then evaluate this. Nine times one-tenth is nine over 10, and a half multiplied by two is one. So we get nine over 10 minus one, which is negative one-tenth, which is our final answer.

Therefore, we were able to show the limit as π‘₯ approaches zero of nine π‘₯ divided by the sin of 10π‘₯ minus the tan of two π‘₯ divided by two π‘₯ is negative one over 10.

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