Determine the following limits.
lim θ→∞ cos θ / θ

Question

Determine the following limits.

lim θ→∞ cos θ / θ²

Accepted Answer

Welcome back, everyone. In this problem, we want to find the limit at infinity for the limit of 3 x divided by X to the 6th as X approaches infinity. A says it's 0, B negative infinity, C infinity, and D says it's 3. Now how are we going to figure this limit out? Well, first, let's start to think about the nature of the sign function. What do we know? Well, recall, OK, that the sine function is always between -1 and 1, OK? Those are the boundaries for the sine function. Here, since we're looking at 3 sin x, if we multiply our boundary by 3, then this implies that -3 is going to be less than or equal to 3 X X, which is going to be equal to, sorry, which is going to be less than or equal to 3, OK? And if we divide our bound by X to the 6th, that means. The 3 divided by x 6, less than R equal to 3 x divided by x 6, which is less than or equal to 3 divided by X to the 6th. Now, why did we write this? Why did we come up with this boundary to help us find the limit of our function at infinity? Well, if we recall the squeeze theorem, OK, by the squeeze theorem. It can help us to find the limit because by the squeeze theorem, it tells us that if F of X, OK. is less than or equal to geofix, which is less than or equal to HFX. In other words, it's just saying that if GeoffIX is between two other functions, FX and HFX, OK. When X is near A, OK. Where it represents a a value, some value, OK. So let me just clean this up here. OK. So when X is near A, except possibly at A, OK. And The limit of FF X as X approaches A. is equal to the limit of H of X as X approaches A, OK, which is equal to some finite limit L. Then That tells us then that the limit, OK, as X approaches A of G of X is also going to be equal to L. Now if you notice here, based on the squeeze theorem, 3 x divided by x 6 is squeezed between -3 divided by x 6 and 3 divided by X to the 6th. So if we can find the limits of both of those terms, and if they are equal, then that is also going to be the limit of 3 sin x divided by X to the 6th. Now what are the, are those limits going to be? Well, starting with the limit as X approaches infinity of -3 divided by X to the 6th, in this limit, X as X approaches infinity to the 6 becomes arbitrarily large, which means that this function is going to tend to zero. Likewise, our limit. As X approaches infinity of 8 X 3 divided by X to the 6th, will also be equal to 0. Since both of them are the same, OK, this tells us then. That the limit I can just make some space here. The limit as x approaches infinity of -3 divided by X to the 6th is equal to the limit as X approaches infinity of 3 divided by X to the 6th, which is 0. Since both of these limits are the same, we can conclude then that the limit as X approaches infinity of 3 sin x. Divided by X 6 will also be equal to 0. Therefore, A is the correct answer. Thanks for watching everyone. I hope this video helped.

Determine the following limits.
lim θ→∞ cos θ / θ²

Key Concepts

Limits

Trigonometric Functions

Dominance of Growth Rates

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