Video: Using the Squeeze Theorem to Evaluate a Limit Involving a Trigonometric Function

Use the squeeze theorem to evaluate lim_(πœƒ β†’ 0) 2πœƒΒ² cos (1/πœƒ).

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Video Transcript

Use the squeeze theorem to evaluate the limit of two πœƒ squared cos of one over πœƒ as πœƒ approaches zero.

Let’s recall the squeeze theorem. The squeeze theorem says that if there exist functions 𝑓 of π‘₯, 𝑔 of π‘₯, and β„Ž of π‘₯ such that 𝑔 of π‘₯ is greater than or equal to 𝑓 of π‘₯ but less than or equal to β„Ž of π‘₯ when π‘₯ is near π‘Ž, except possibly at π‘Ž. And the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ equals the limit as π‘₯ approaches π‘Ž of β„Ž of π‘₯ equals 𝐿. Then the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯ equals 𝐿.

In order to use the squeeze theorem, we need to find a function 𝑔 of π‘₯ that we can bound from above and below by functions β„Ž of π‘₯ and 𝑓 of π‘₯, respectively. The limit we are asked to evaluate in the question contains the function cos of one over πœƒ. Note that the value cos of π‘₯ is always less than or equal to one and greater than or equal to negative one for any of real number π‘₯. And so the function cos of one over πœƒ is bounded from above by the constant function one and bounded from below by the constant function negative one.

Now let’s multiply these inequalities by two πœƒ squared. Doing so, we obtain two πœƒ squared cos of one over πœƒ, which is the function whose limit we’re asked to evaluate as πœƒ approaches zero, is less than or equal to two πœƒ squared and greater than or equal to negative two πœƒ squared. Note that these bounds hold for any real number πœƒ near zero apart from πœƒ equals zero, as one over πœƒ is not defined when πœƒ is equal to zero.

As a result, letting the function 𝑔 equal to πœƒ squared cos of one over πœƒ, the function 𝑓 equal negative two πœƒ squared, the function β„Ž equal two πœƒ squared, and π‘Ž equal zero in the squeeze theorem. We find that if the limit of negative two πœƒ squared as πœƒ approaches zero equals the limit of two πœƒ squared as πœƒ approaches zero equals 𝐿. Then the limit of two πœƒ squared cos of one over πœƒ as πœƒ approaches zero also equals 𝐿.

Let’s see if we can evaluate the limit as πœƒ approaches zero of negative two πœƒ squared and the limit as πœƒ approaches zero of two πœƒ squared. Firstly, note that we can factor the multiplicative constant out of a limit. Therefore, the limit of negative two πœƒ squared as πœƒ approaches zero is equal to negative two multiplied by the limit of πœƒ squared as πœƒ approaches zero. For the same reason, the limit of two πœƒ squared as πœƒ approaches zero is equal to two multiplied by the limit of πœƒ squared as πœƒ approaches zero.

Now let’s evaluate the limit as πœƒ approaches zero of πœƒ squared. In order to do this, we will use the fact that the limit of π‘₯ to the power of 𝑛 as π‘₯ approaches π‘Ž is equal to π‘Ž to the power of 𝑛 for all positive integers 𝑛. Substituting π‘Ž equals zero and 𝑛 equals two into this result, we find that the limit of πœƒ squared as πœƒ approaches zero is equal to zero squared, which is just zero.

We therefore find that the limit of negative two πœƒ squared as πœƒ approaches zero is equal to negative two multiplied by zero, which equals zero. And the limit of two πœƒ squared as πœƒ approaches zero is equal to two multiplied by zero, which also equals zero. So the limit of negative two πœƒ squared as πœƒ approaches zero is equal to the limit of two πœƒ squared as πœƒ approaches zero. And they are both equal to zero.

Therefore, by the squeeze theorem, the limit of two πœƒ squared cos of one over πœƒ as πœƒ approaches zero is also equal to zero, which is our final answer.

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