Video: Finding the Limit of a Quotient of Trigonometric and Linear Functions at a Point

Find lim_(π‘₯ β†’ 0) (6 βˆ’ 3π‘₯)/cos 5π‘₯.

03:02

Video Transcript

Find the limit as π‘₯ approaches zero of six minus three π‘₯ all divided by the cos of five π‘₯.

The question is asking us to evaluate the limit as π‘₯ approaches zero of the quotient of two functions. The first thing we should do when we’re asked to evaluate a limit like this is ask the question, can we evaluate this by using direct substitution?

First, we see we’re asked to evaluate the limit of a quotient. And we know if 𝑓 of π‘₯ is a quotient of two functions, 𝑔 of π‘₯ and β„Ž of π‘₯. And we can evaluate both 𝑔 and β„Ž by direct substitution when π‘₯ is approaching π‘Ž. And in particular, we want our denominator β„Ž evaluated at π‘Ž to not be equal to zero. Then we can evaluate the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ by direct substitution.

In our case, our numerator 𝑔 of π‘₯ is equal to six minus three π‘₯, and our denominator β„Ž of π‘₯ is equal to the cos of five π‘₯. We need to show that both 𝑔 and β„Ž of π‘₯ can be evaluated by direct substitution as π‘₯ approaches π‘Ž. And we also need that β„Ž evaluated at π‘Ž is not equal to zero. If we can show that all of this is true, then we can evaluate our original limit by direct substitution.

Let’s start with our numerator 𝑔 of π‘₯. We see that 𝑔 of π‘₯, which is equal to six minus three π‘₯, is a linear function, which itself means it’s a polynomial. And we know that all polynomials can be evaluated by direct substitution for all real values π‘Ž. So our numerator 𝑔 of π‘₯ can be evaluated by direct substitution as π‘₯ is approaching zero.

Next, we need to show that our function β„Ž of π‘₯ can be evaluated by direct substitution. And our function β„Ž of π‘₯ is equal to the cos of five π‘₯, which is a trigonometric function. And we know that all trigonometric functions can be evaluated by direct substitution as long as π‘₯ is approaching some value within their domain. In our case, we’re taking the limit as π‘₯ approaches zero. So we want our value of π‘Ž to be equal to zero. And we know the cos of five π‘₯ is defined for all real values of π‘₯. So in particular, zero is in the domain of this function. So if we set π‘Ž equal to zero, we’ve shown our denominator β„Ž of π‘₯ can be evaluated by direct substitution as π‘₯ approaches zero.

The last thing we need is that β„Ž evaluated at zero is not equal to zero. And we can evaluate this directly. β„Ž evaluated at zero is the cos of five times zero, which is equal to the cos of zero, which we know is equal to one. So β„Ž evaluated at zero is not equal to zero. This tells us we can evaluate our limit by using direct substitution. So we’ll substitute π‘₯ is equal to zero into our function 𝑓 of π‘₯, which is 𝑔 of π‘₯ divided by β„Ž of π‘₯. This gives us six minus three times zero divided by the cos of five times zero.

And we can just evaluate this expression. Our numerator of six minus three times zero is just six minus zero. And the cos of five times zero is the cos of zero, which is equal to one. So we get six divided by one, which is just equal to six. We’ve shown the limit as π‘₯ approaches zero of six minus three π‘₯ divided by the cos of five π‘₯ is equal to six.

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