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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,

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