Video Transcript
Knowing that the squeeze theorem applies when the limit is taken at ∞, determine the limit as 𝑥 approaches ∞ of sin of three 𝑥 over 𝑥.
Let’s recall the squeeze theorem. The squeeze theorem says that if there exists 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.
The statement of the question addressed in this video tells us that we can assume that the squeeze theorem applies when the limit is taken at ∞. So we can interpret the squeeze theorem we know as follows. If 𝑓 of 𝑥, 𝑔 of 𝑥, and ℎ of 𝑥 are functions 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 approaches ∞. And the limit as 𝑥 approaches ∞ of 𝑓 of 𝑥 equals the limit as 𝑥 approaches ∞ of ℎ of 𝑥 equals 𝐿. Then the limit as 𝑥 approaches ∞ 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, when 𝑥 approaches ∞. The limit we are asked to determine in the question contains the function sin of three 𝑥. Note that the value sin of 𝑥 is always less than or equal to one and greater than or equal to negative one for any real number 𝑥. So replacing 𝑥 by three 𝑥, we obtain that the function sin of three 𝑥 is bounded from above by the constant function one and bounded from below by the constant function negative one.
Now, let’s divide these inequalities by 𝑥. Doing so, we obtain that sin of three 𝑥 over 𝑥, which is the function whose limit we are asked to determine as 𝑥 approaches ∞. Is less than or equal to one over 𝑥 and greater than or equal to negative one over 𝑥. Note that these bounds hold for all real numbers 𝑥, apart from 𝑥 equals zero as the function one over 𝑥 is not defined when 𝑥 equals zero. In particular, these bounds hold when 𝑥 is large, i.e., approaches ∞.
As a result, letting the function 𝑔 equal sin of three 𝑥 over 𝑥, the function 𝑓 equal negative one over 𝑥, and the function ℎ equal one over 𝑥. In the version of the squeeze theorem, when the limit is taken at ∞, we find that if the limit as 𝑥 approaches ∞ of negative one over 𝑥 equals the limit as 𝑥 approaches ∞ of one over 𝑥 equals 𝐿. Then the limit as 𝑥 approaches ∞ of sin of three 𝑥 over 𝑥 equals 𝐿. Recall the standard result, the limit as 𝑥 approaches ∞ of one over 𝑥 equals zero.
Now, in order to find the limit as 𝑥 approaches ∞ of negative one over 𝑥, we use the property that we can factor multiplicative constants out of a limit. Using that property, we obtain that the limit as 𝑥 approaches ∞ of negative one over 𝑥 is equal to the negative of the limit as 𝑥 approaches ∞ of one over 𝑥, which is just zero. So the limit as 𝑥 approaches ∞ of negative one over 𝑥 is equal to the limit as 𝑥 approaches ∞ of one over 𝑥. And they are both equal to zero.
Therefore, by the squeeze theorem, the limit as 𝑥 approaches ∞ of sin of three 𝑥 over 𝑥 is also equal to zero. This is our final answer.
