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Section 2-5 : Computing Limits

14. Given that \(7x \le f\left( x \right) \le 3{x^2} + 2\) for all \(x\) determine the value of \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\).

Hint : Recall the Squeeze Theorem.
Show Solution

This problem is set up to use the Squeeze Theorem. First, we already know that \(f\left( x \right)\) is always between two other functions. Now all that we need to do is verify that the two “outer” functions have the same limit at \(x = 2\) and if they do we can use the Squeeze Theorem to get the answer.

\[\mathop {\lim }\limits_{x \to 2} 7x = 14\hspace{0.75in}\mathop {\lim }\limits_{x \to 2} \left( {3{x^2} + 2} \right) = 14\]

So, we have,

\[\mathop {\lim }\limits_{x \to 2} 7x = \mathop {\lim }\limits_{x \to 2} \left( {3{x^2} + 2} \right) = 14\]

and so by the Squeeze Theorem we must also have,

\[\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 14\]