Question Video: Using the Squeeze Theorem to Evaluate a Limit | Nagwa Question Video: Using the Squeeze Theorem to Evaluate a Limit | Nagwa

# Question Video: Using the Squeeze Theorem to Evaluate a Limit Mathematics • Higher Education

Given that β3π₯Β² + 9π₯ β 5 β€ π(π₯) β€ π₯Β² β 3π₯ + 4, find lim_(π₯ β 3/2) π(π₯).

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### Video Transcript

Given that π 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 π of π₯.

In this question, we are given a function π 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 π 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 π of π₯, π of π₯, and β of π₯, such that π of π₯ is greater than or equal to π of π₯ but less than or equal to β of π₯. When π₯ is near π, except possibly at π, and the limit as π₯ approaches π of π of π₯ equals the limit as π₯ approaches π of β of π₯ equals πΏ. Then the limit as π₯ approaches π of π of π₯ equals πΏ.

In order to use the squeeze theorem, we need to have a function π of π₯ that is bounded from above and below by functions β of π₯ and π of π₯, respectively. We are told that the function π 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 π of π₯ equal negative three π₯ squared plus nine π₯ minus five, β of π₯ equal π₯ squared minus three π₯ plus four, and π 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 πΏ. Then the limit as π₯ approaches three over two of π of π₯ equals πΏ. Letβ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 π of π₯ also equals seven over four. Hence, we have shown, using the squeeze theorem, that if the function π 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 π of π₯ equals seven over four.

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