Video: Finding the Limit of a Product of Linear and Trigonometric Functions

Determine lim_(π‘₯ β†’ πœ‹/3) 7π‘₯ cos π‘₯.

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Video Transcript

Determine the limit as π‘₯ approaches πœ‹ by three of seven π‘₯ times the cos of π‘₯.

We see the question wants us to evaluate the limit as π‘₯ approaches πœ‹ by three of seven π‘₯ multiplied by the cos of π‘₯. We can see that this is the limit of a linear function multiplied by a trigonometric function. We want to evaluate this by direct substitution, but let’s check that we can do that in this case.

First, we recall that we can evaluate the limit of any polynomial by using direct substitution. And since seven π‘₯ is a linear function which is also a polynomial, this means we can in fact evaluate the limit of seven π‘₯ by direct substitution. Next, we also know that we can evaluate the limit of a trigonometric function by direct substitution as long as we’re taking the limit as π‘₯ approaches some value within this trigonometric function’s domain.

The trigonometric function in this question is the cos of π‘₯, which we know is defined for all values of π‘₯. So, this means we can evaluate the limit as π‘₯ approaches πœ‹ by three of the cos of π‘₯ by direct substitution. But we don’t want to evaluate the limit of each of these functions separately by using direct substitution. We want to evaluate the limit of their product by using direct substitution.

At this stage, there are now two different ways we can evaluate this limit. First, we could use the fact that the limit of a product is equal to the product of the limits. Using this, we get the limit as π‘₯ approaches πœ‹ by three of seven π‘₯ times the cos of π‘₯ is equal to the limit as π‘₯ approaches πœ‹ by three of seven π‘₯ times the limit as π‘₯ approaches πœ‹ by three of the cos of π‘₯.

We see our first limit is the limit of a polynomial and our second limit is the limit of a trigonometric function. So, we can evaluate each of these limits by direct substitution. Substituting π‘₯ is equal to πœ‹ by three, we get seven times πœ‹ by three multiplied by the cos of πœ‹ by three. And we know that the cos of πœ‹ by three is equal to one-half. So, we can calculate this expression to give us seven πœ‹ by six.

However, we don’t actually need to split this limit into a product of two limits. We know if we can evaluate the limit of two functions 𝑓 of π‘₯ and 𝑔 of π‘₯ by direct substitution as π‘₯ approaches some value of π‘Ž, then we can actually also evaluate their product by using direct substitution. So, to use this on the limit as π‘₯ approaches πœ‹ by three of seven π‘₯ times the cos of π‘₯, just as we did before, we know we can evaluate the limit of seven π‘₯ and the limit of the cos of π‘₯ as π‘₯ approaches πœ‹ by three by direct substitution. Because seven π‘₯ is a polynomial and the cos of π‘₯ is a trigonometric function.

So, we can attempt to evaluate this expression by just substituting π‘₯ is equal to πœ‹ by three. So, we substitute π‘₯ is equal to πœ‹ by three, and we get seven times πœ‹ by three multiplied by the cos of πœ‹ by three. And again, we can evaluate this expression to give us seven πœ‹ by six. So, we’ve shown by using direct substitution the limit as π‘₯ approaches πœ‹ by three of seven π‘₯ times the cos of π‘₯ is equal to seven πœ‹ by six.

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