Question Video: Finding Limits Involving Trigonometric Functions

Find lim_(π‘₯ β†’ 0) 9 cot π‘₯ sin 4π‘₯.

03:52

Video Transcript

Find the limit as π‘₯ approaches zero of nine times the cot of π‘₯ times the sin of four π‘₯.

The question is asking us to find the limit of a combination of trigonometric functions. And we might be tempted to use direct substitution. However, since our limit is as π‘₯ is approaching zero, we see that the cot of zero does not exist. So we’re going to need to manipulate this expression into one which we can use direct substitution. Or we’re going to need to use one of our many limit laws for trigonometric functions. We start by recalling one of our trigonometric identities. The cot of π‘₯ is equivalent to one divided by the tan of π‘₯, which is the same as saying the cos of π‘₯ divided by the sin of π‘₯.

So by using our trigonometric identity, we have our limit is the same as the limit as π‘₯ approaches zero of nine times the cos of π‘₯ divided by the sin of π‘₯ multiplied by the sin of four π‘₯. And we see this is the limit of trigonometric functions. So we could try using direct substitution. Substituting π‘₯ is equal to zero into our limit would give us nine times the cos of zero divided by the sin of zero multiplied by the sin of four times zero. However, the sin of zero is equal to zero. So if we evaluated this expression, we would get the indeterminate form zero divided by zero. So we still need to do more manipulation to evaluate our limit.

To help us evaluate this limit, we recall the double-angle formula for sin, which tells us the sin of two πœƒ is equivalent to two times the sin of πœƒ times the cos of πœƒ. However, we want to use this on the sin of four π‘₯. So we’ll set πœƒ equal to two π‘₯. This gives us that the sin of four π‘₯ is equivalent to two times the sin of two π‘₯ times the cos of two π‘₯. So by using our double-angle formula, we have that our limit is equal to the limit as π‘₯ approaches zero of nine times the cos of π‘₯ divided by the sin of π‘₯ multiplied by two sin of two π‘₯ cos of two π‘₯. However, we still can’t use direct substitution. We see our limit is as π‘₯ is approaching zero. In our denominator, we have the sin of zero, which is equal to zero. And on our numerator, we have the sin of two times zero, which is also zero.

This would again leave us with the indeterminate form zero divided by zero. However, if we were to use the double-angle formula on the sin of two π‘₯ in our numerator, we would get a factor of the sin of π‘₯ in our numerator, which we could cancel with the sin of π‘₯ in our denominator. This means we would no longer get an indeterminate form when trying direct substitution. So by using the fact that the sin of two π‘₯ is equivalent to two times the sin of π‘₯ cos of π‘₯, we can rewrite our limit as the limit as π‘₯ approaches zero of nine times the cos of π‘₯ divided by the sin of π‘₯ times two times two sin of π‘₯ cos of π‘₯ all multiplied by the cos of two π‘₯. And we can now simplify this limit. We can cancel the shared factor of the sin of π‘₯ in our numerator and our denominator. And we also have that nine multiplied by two multiplied by two is equal to 36.

Finally, we can combine both of our factors of the cos of π‘₯ to give us the cos squared of π‘₯. So we’ve shown that our limit is equal to the limit as π‘₯ approaches zero of 36 cos squared of π‘₯ times the cos of two π‘₯. And now that we’ve removed the factor of sin of π‘₯ from our denominator, we can use direct substitution. Substituting π‘₯ is equal to zero gives us 36 times the cos squared of zero times the cos of two times zero. And we know the cos of zero is equal to one and so the cos of zero squared is also equal to one. Similarly, the cos of two times zero is equal to one. This gives us that our limit is equal to 36. Therefore, we’ve shown the limit as π‘₯ approaches zero of nine times the cot of π‘₯ times the sin of four π‘₯ is equal to 36.

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