1. Nov 7, 2008

### xiaochobitz

1. The problem statement, all variables and given/known data

2. Relevant equations
$$\lim(\frac{(x^2)-2x-3}{(x^2)+5x+6})$$
$$x->(-2)$$

3. The attempt at a solution

anyone can help out on this one?

2. Nov 7, 2008

### Dick

Factor the quadratics. Which factor is going to zero? Can you cancel it?

3. Nov 7, 2008

### xiaochobitz

er by factoring it i actually get

$$\lim(\frac{(x-3)(x+1)}{(x+3)(x+2)})$$
$$x->(-2)$$

am stuck from here on~

4. Nov 7, 2008

### Dick

Ok, it doesn't cancel. My mistake. But the limit of the numerator is 5 and the limit of the denominator is 0. What does that tell you about the limit of the quotient?

5. Nov 7, 2008

### xiaochobitz

tends to infinity?

6. Nov 7, 2008

### Dick

It's worse than that. It goes to +infinity from one side and -infinity from the other. I would just say 'doesn't exist'.

7. Nov 7, 2008

### xiaochobitz

so there is a problem in the question? haha~

8. Nov 7, 2008

### Dick

Not necessarily. They can ask you to find a limit and expect you to determine that it doesn't exist. That's a possible correct answer.

9. Nov 7, 2008

### xiaochobitz

hmm icic~ troublesome question that got me stuck for so long haha~

anyways, appreciate the help given, thanks so much~

10. Nov 7, 2008

### whitay

L'Hopitals?

( 2x - 2 ) / ( 2x + 5 ) -> -2 is -6

Just realised L'Hop wont work cause it isn't 0/0. Sorry.

or

if i divide through by the highest power I get -5/12.

I don't have anything to graph with, but plot it and see.

Last edited: Nov 7, 2008
11. Nov 7, 2008

### Dick

l'Hopital, NO. And you know why. Dividing through by the highest power doesn't help. x doesn't go to infinity, you can't say the other terms go to zero. Plotting it will show there is no limit. As you should guess, the form is 5/0.

12. Nov 8, 2008

### HallsofIvy

Staff Emeritus
It the numerator and denomenator separately go to 0, you can use L'Hopital. If the denominator goes to 0 and the numerator doesn't, the limit does not exist.