Using the squeeze theorem to solve a limit

[LOLItsAJ]

Homework Statement

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.

Homework Equations

Lim (x^2-9)(x-3/|x-3|)
x\rightarrow3

The Attempt at a Solution

I've tried so many attempts at this problem, posting all my work would be massive.

 

[SammyS]

Homework Statement

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.

Homework Equations

Lim (x^2-9)(x-3/|x-3|)
x\rightarrow3

The Attempt at a Solution

I've tried so many attempts at this problem, posting all my work would be massive.

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)\,?

 

[LOLItsAJ]

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

 

[SammyS]

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|}&lt; 8|x-3| for 2 < x < 4 ?