Video: Finding Limits Involving Trigonometric Functions

Find lim_(π‘₯ β†’ πœ‹/2) (2 βˆ’ 2 sin π‘₯)/(4π‘₯ βˆ’ 2πœ‹).

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Video Transcript

Find the limit as π‘₯ goes to πœ‹ over two of two minus two sin of π‘₯ over four π‘₯ minus two πœ‹.

First, we try solving it by direct substitution. However, we’ll reach zero over zero, which is undefined. So we must find this limit by another means. Let’s start by cancelling a factor of two in both the numerator and denominator of the fraction. Now let’s consider some of the rules we know. When considering the first rule here, we notice that the value inside of the sine must be the same as the value in the denominator of the function. In our limit, we have sin of π‘₯ in the numerator. However, in the denominator, we have two π‘₯ minus πœ‹. And these two things are not equal. Therefore, we cannot use this first rule.

In order to use the second rule, we require a cos π‘₯ in the numerator. However, in our limit, we currently have a sine. Here we’ll be using an identity in order to change the sine to a cosine. We have that sin of π‘₯ is equal to cos of π‘₯ minus πœ‹ by two. And we can substitute this into our limit. Factorizing the denominator of the fraction here, we can see that this is now of a very similar form to the rule which we know.

At this stage, we need to perform a substitution. We will substitute in 𝑒 is equal to π‘₯ minus πœ‹ by two. However, we first need to consider what will happen to our limit. So that’s π‘₯ approaching πœ‹ by two. Well, we’ll simply consider what happens to the value of 𝑒 as π‘₯ approaches πœ‹ by two. Our value of 𝑒 will approach πœ‹ by two minus πœ‹ by two, which is simply zero. And so now we’re ready to substitute 𝑒 is equal to π‘₯ minus πœ‹ by two into our limit. We obtain the limit as 𝑒 approaches zero of one minus cos of 𝑒 over two 𝑒. We can factor out the half. And now we notice that our limit is identical to our rule. And so, therefore, it must equal zero. And this gives us a solution of zero.

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