Video: Using the Squeeze Theorem to Evaluate a Limit

Given that βˆ’3π‘₯Β² + 9π‘₯ βˆ’ 5 ≀ 𝑓(π‘₯) ≀ π‘₯Β² βˆ’ 3π‘₯ + 4, find lim_(π‘₯ β†’ 3/2) 𝑓(π‘₯).

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Video Transcript

Given that 𝑓 of π‘₯ is greater than or equal to negative three π‘₯ squared plus nine π‘₯ minus five and less than or equal to π‘₯ squared minus three π‘₯ plus four, find the limit as π‘₯ approaches three over two of 𝑓 of π‘₯.

In this question, we are given a function 𝑓 of π‘₯ and told that it is bounded from above by the function π‘₯ squared minus three π‘₯ plus four and bounded from below by the function negative three π‘₯ squared plus nine π‘₯ minus five. We are asked to find the limit as π‘₯ approaches three over two of 𝑓 of π‘₯.

The fact that we are asked to find the limit of a function, which is bounded between two other functions, reminds us of the squeeze theorem. The squeeze theorem says that if there exist functions 𝑔 of π‘₯, 𝑓 of π‘₯, and β„Ž of π‘₯, such that 𝑓 of π‘₯ is greater than or equal to 𝑔 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 𝑔 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 𝑓 of π‘₯ equals 𝐿.

In order to use the squeeze theorem, we need to have a function 𝑓 of π‘₯ that is bounded from above and below by functions β„Ž of π‘₯ and 𝑔 of π‘₯, respectively. We are told that the function 𝑓 of π‘₯ whose limit we are asked to find as π‘₯ approaches three over two is bounded from above by the function π‘₯ squared minus three π‘₯ plus four. And bounded from below by the function negative three π‘₯ squared plus nine π‘₯ minus five. Since the question does not state any particular π‘₯-values for which these bounds hold, we can assume that these bounds hold for all real numbers π‘₯ or at least for values of π‘₯ near three over two.

So letting 𝑔 of π‘₯ equal negative three π‘₯ squared plus nine π‘₯ minus five, β„Ž of π‘₯ equal π‘₯ squared minus three π‘₯ plus four, and π‘Ž equal three over two. In the squeeze theorem, we deduce that if the limit as π‘₯ approaches three over two of negative three π‘₯ squared plus nine π‘₯ minus five equals the limit as π‘₯ approaches three over two of π‘₯ squared minus three π‘₯ plus four equals 𝐿. Then the limit as π‘₯ approaches three over two of 𝑓 of π‘₯ equals 𝐿. Let’s evaluate the limit as π‘₯ approaches three over two of negative three π‘₯ squared plus nine π‘₯ minus five and the limit as π‘₯ approaches three over two of π‘₯ squared minus three π‘₯ plus four.

Since negative three π‘₯ squared plus nine π‘₯ minus five and π‘₯ squared minus three π‘₯ plus four are polynomial expressions and the value three over two is a real number. We can evaluate the limits using direct substitution. Doing so, we obtain that the limit as π‘₯ approaches three over two of negative three π‘₯ squared plus nine π‘₯ minus five equals negative three timesed by three over two squared plus nine times three over two minus five, which equals seven over four. We obtain that the limit as π‘₯ approaches three over two of π‘₯ squared minus three π‘₯ plus four equals three over two squared minus three times three over two plus four, which also equals seven over four.

So, we have that the limit as π‘₯ approaches three over two of negative three π‘₯ squared plus nine π‘₯ minus five equals the limit as π‘₯ approaches three over two of π‘₯ squared minus three π‘₯ plus four. And they are both equal to seven over four. So by the squeeze theorem, the limit as π‘₯ approaches three over two of 𝑓 of π‘₯ also equals seven over four. Hence, we have shown, using the squeeze theorem, that if the function 𝑓 of π‘₯ is bounded from above by the function π‘₯ squared minus three π‘₯ plus four and bounded from below by the function negative three π‘₯ squared plus nine π‘₯ minus five. Then the limit as π‘₯ approaches three over two of 𝑓 of π‘₯ equals seven over four.

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