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Section 2.5 : Computing Limits

14. Given that \(x^{3}-6x^{2}+12x-3 \le f\left( x \right) \le x^{2}-4x+9\) for \(x \le 3\) determine the value of \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\).

Hint : Recall the Squeeze Theorem.
Show Solution

First, given that we know \(f\left( x \right)\) is always between two other functions for \( x\le 3\) and \(x = 2\) is in this range this problem is set up for the Squeeze Theorem. 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.

\[\mathop {\lim }\limits_{x \to 2} \left(x^{3}-6x^{2}+12x-3\right) = 5\hspace{0.75in}\mathop {\lim }\limits_{x \to 2} \left( x^{2}-4x+9 \right) = 5\]

So, we have,

\[\mathop {\lim }\limits_{x \to 2} \left(x^{3}-6x^{2}+12x-3\right) = \mathop {\lim }\limits_{x \to 2} \left( x^{2}-4x+9\right) = 5\]

and so by the Squeeze Theorem we must also have,

\[\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 5\]