# Using the squeeze theorem to solve a limit

1. Sep 2, 2011

### LOLItsAJ

1. The problem statement, all variables and given/known data
I have a limit, which when substitution is used the indeterminate form 0/0 occurs. I've been asked to solve the limit by using the squeeze theorem. I really have no idea where to take this.

2. Relevant equations
Lim (x$^2$-9)(x-3$/$|x-3|)
x$\rightarrow$3

3. The attempt at a solution
I've tried so many attempts at this problem, posting all my work would be massive.

2. Sep 2, 2011

### SammyS

Staff Emeritus
Are you sure the problem is what you state, which is equivalent to $\displaystyle \lim_{x\to3}\left((x^2-9)\left(x-\frac{3}{|x-3|}\right)\right)\,,$ or is it $\displaystyle \lim_{x\to3}\left((x^2-9)\frac{x-3}{|x-3|}\right)\,?$

3. Sep 2, 2011

### LOLItsAJ

It's your second one. I'm a bit confused by the format used on the forum.

4. Sep 2, 2011

### SammyS

Staff Emeritus
It appears to me that simplifying the expression using algebra works better than using the squeeze theorem, but ...

Certainly, $\displaystyle (x^2-9)\frac{x-3}{|x-3|}\ge0$ for x > 0.

Can you show that $\displaystyle (x^2-9)\frac{x-3}{|x-3|}< 8|x-3|$ for 2 < x < 4 ?