Using limθ→0 sin θ / θ = 1

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Question

Using limθ→0 sin θ / θ = 1

Find the limits in Exercises 23–46.

limx→0 (1 − cos 3x) / 2x

Accepted Answer

Welcome back, everyone. Find the limit of the function F of X equals 1 minus cosine of 6 x divided by 4x access X approaches 0. We're given 4 answer choices A says 0, B infinity, C 1, and D does not exist. So let's write down the limit. Limit as X approaches 0 of 1 minus cosine of 6 X divided by 4 X. We can simply attempt direct substitution, assuming that our function is continuous, which gives us 1 minus cosine of 0 divided by 4 multiplied by 0. And this gives us an indeterminate form, 0 divided by 0, right, division by 0 is undefined, and if we have 0 divided by 0, well, essentially this means that we have to rewrite our limit in a different form and use a different way to solve for it. So, what we're going to do in this problem is simply use the definition, limited as X approaches 0 of sin of x divided by X is equal to 1. Now the main problem is that we have cosine instead of sign, but we can rewrite cosine in terms of sine as follows using the half angle formula. Let's recall that 2 sine squared of alpha divided by 2 is equal to 1 minus cosine of alpha. In this context, we have 1 minus cosine of 6 X. so this is going to be equal to 2 squared off half of 6 X which is 3 X, right? So we can essentially say that our limit becomes limit as x approaches 0 of 2 sin square of 3 x divided by. 4 acts Now, what we can do is simply perform simplification. We can divide both sides by 2, right? So this gives us sign squared in the numerator and 2 X in the denominator. Now another problem is that we have sin squared. So what we can do is simply use the properties of limits. We can rewrite this sin squared of 3 x as sin of 3 x multiplied by sin of 3 X, right? So let's isolate to limits. One of them is going to be limit as x approaches 0 of. Sign of 3 X divided by 2, we can also incorporate that constant in the denominator multiplied by a limit. As X approaches 0 of sin of 3 x divided by X. If we go backwards, we can see that this product results in sin squared of 3 X divided by 2 Xs, right? So we have applied the properties of limits. And now, we already We can start evaluating it right now for the second limit, we have limit as x approaches 0 of sin of 3 x divided by X. Let's recall our definition. We want to make sure that our angle represents the same value that we have in the denominator. It must be X and X. In this case, we have 3 X for the angle, but we only have X for the denominator. So what we can do is simply multiply and divide both sides of the fraction by 3, right, because this gives us 3 X which corresponds to the angle of sine. And now let's continue by applying the properties of limits. We get limit as X approaches 0, we can factor out the constant. It's 12. Of sign of 3 x multiplied by. Now, factoring out the constant, we can take 3 out. Limet X approaches 0 of sign of 3 X divided by 3 X. Now the very last issue that we have is that X approaches 0 and we have 3 X for the angle of sign, right? So if X approaches 0, we can also state that 3 X will also approach 0. We're simply multiplying both sides by 3, right? So we can clearly see that now we have a perfect definition. That we previously proposed limit as 3 x approaches 0 of sign of 3 x divided by 3 X is going to be equal to 1, right? So this limit is going to be equal to 1, and now we can perform the calculations. We get 1/2 multiplied by a limit of sin of 3 X as X approaches 0, that's simply 0. We can use direct substitution, sign of 0 is 0. And then we're multiplying by 3, which is multiplied by 1. Overall, the answer is 0, which corresponds to the answer choice A. Thank you for watching.

Using limθ→0 sin θ / θ = 1

Find the limits in Exercises 23–46.

limx→0 (1 − cos 3x) / 2x

Key Concepts

Limit Definition

Trigonometric Limits

L'Hôpital's Rule

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