I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
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 \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\).
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,
\[\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 5\]