# Need some help with lim

1. Mar 30, 2005

### bayan

Hey every one!

I have been doing some lim questions and I came across this one.

Evaluate: lim x-->3 (2x^2-x-15/3x^2-13x+12)

I have Factorised It and got

lim x-->3 ( (x-8) (x+7) / (-3x+4) (-x+3) )

How do I go about the rest?Have I used the right stuff?

Sorru for the mess. I don't realy know how to use the latex

2. Mar 30, 2005

### matt grime

You've not factored the top properly. If that were the correct factorization then the limit wouldn't exist. However, 2x^2 - x -15 evaluates to zero at x=3, so x-3 must be one of the factors., which will cancel with the factor or x-3 on the bottom.

3. Mar 30, 2005

### bayan

thanx!

I see my problem now :)

4. Mar 30, 2005

### bayan

After Factorising I would just substitude the x-->3 into the equation wouldnt I?

Or do I have to use the f(x+h)-f(x)/h? (not too sure about formula too:( I'm feeling kinda sleeeeepppy)

5. Mar 30, 2005

### dextercioby

Nope,u'd simplify through the common monom (x-3) and then take the limit...

Daniel.

6. Mar 30, 2005

### bayan

I think I need more explenation.

After canceling (x-3) from top and bottom do I just change the value of x=3? That is what I currently have done! And got 6+5/9-4=11/5=2 1/5

7. Mar 30, 2005

### dextercioby

That's right.The limit in that case means plugging the value "x=3" in the remaining fraction.

Daniel.

8. Mar 30, 2005

### bayan

Thanx.

Sorry to bother you, But I really apriciate the help.

9. Mar 30, 2005

### Daminc

Factorising 2X^2-X-15 gives
(2X+5)(X-3)

Therefore:
X=3 or
X=-5/2

10. Mar 30, 2005

### dextercioby

We don't need the second value (-5/2).Jus that "3" was important,because it got simplified with the one in the denominator.

Daniel.