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

Use the squeeze theorem to evaluate lim_(𝜃 → 0) 2𝜃² cos (1/𝜃).

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### Video Transcript

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

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

In order to use the squeeze theorem, we need to find a function 𝑔 of 𝑥 that we can bound from above and below by functions ℎ of 𝑥 and 𝑓 of 𝑥, respectively. The limit we are asked to evaluate in the question contains the function cos of one over 𝜃. Note that the value cos of 𝑥 is always less than or equal to one and greater than or equal to negative one for any of real number 𝑥. And so the function cos of one over 𝜃 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 𝜃 squared. Doing so, we obtain two 𝜃 squared cos of one over 𝜃, which is the function whose limit we’re asked to evaluate as 𝜃 approaches zero, is less than or equal to two 𝜃 squared and greater than or equal to negative two 𝜃 squared. Note that these bounds hold for any real number 𝜃 near zero apart from 𝜃 equals zero, as one over 𝜃 is not defined when 𝜃 is equal to zero.

As a result, letting the function 𝑔 equal to 𝜃 squared cos of one over 𝜃, the function 𝑓 equal negative two 𝜃 squared, the function ℎ equal two 𝜃 squared, and 𝑎 equal zero in the squeeze theorem. We find that if the limit of negative two 𝜃 squared as 𝜃 approaches zero equals the limit of two 𝜃 squared as 𝜃 approaches zero equals 𝐿. Then the limit of two 𝜃 squared cos of one over 𝜃 as 𝜃 approaches zero also equals 𝐿.

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

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

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

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