Using the squeeze theorem on a function with absolute value and a polynomial

absolute valuecalculuslimits

I have a problem that requires me to use the squeeze theorem to evaluate a limit, even though it is solvable with algebraic manipulation and direct substitution. I understand how to do the latter here, but not how to find the bounding functions.

Here is the limit:
$$ \lim_{x\to 3}(x^2 -9) \frac{x-3}{\lvert x – 3 \rvert} $$

I understand how to evaluate it and find 0 using algebraic manipulation, but not how to do so using the squeeze theorem. In general, I'm still having some trouble finding bounding functions for the squeeze theorem.

Best Answer

Note that for $x\in(2,4)\setminus\{3\}$, $$\left|(x^2-9)\frac{x-3}{|x-3|}\right|=|x^2-9|=(x+3)(x-3)$$ and for example $$5(x-3)\le (x+3)(x-3)\le 7(x-3)$$

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