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

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.

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 $$5(x-3)\le (x+3)(x-3)\le 7(x-3)$$

Let $$f$$ be the function in the post.

Then $$0 \leq |f(x)|=|x-3||x+3|$$

For $$x \neq 3$$, $$\Big| \frac{x-3}{|x-3|} \Big|=1$$ Thus

$$\lim_{x\to 3} \Big|(x^2 -9) \frac{x-3}{\lvert x - 3 \rvert} \Big| = \lim_{x\to 3} |(x^2 - 9)| = 0$$Which implies $$\lim_{x \to 3} (x^2- 9)=0.$$

Note that $$\lim_{x \to a} |f(x)| = 0$$ implies $$\lim_{x \to a} f(x) = 0$$, since $$-|f(x)| \leq f(x) \leq |f(x)|$$.

Put $$f(x) = x^2 - 9$$ and $$a = 3$$.