Limits for Quadratic Equations

[xiaochobitz]

Homework Statement

Homework Equations

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

The Attempt at a Solution

anyone can help out on this one?

 

[Dick]

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

 

[xiaochobitz]

er by factoring it i actually get

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

am stuck from here on~

 

[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?

 

[xiaochobitz]

tends to infinity?

 

[Dick]

tends to infinity?

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

 

[xiaochobitz]

so there is a problem in the question? haha~

 

[Dick]

so there is a problem in the question? haha~

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.

 

[xiaochobitz]

hmm icic~ troublesome question that got me stuck for so long haha~anyways, appreciate the help given, thanks so much~

 

[whitay]

l'hospital's?

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

Just realized L'Hop won't 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.

 

[Dick]

l'hospital's?

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

Just realized L'Hop won't 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.

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.

 

[HallsofIvy]

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.