Limits for Quadratic Equations

In summary, the conversation discusses finding the limit of the equation (x^2-2x-3)/(x^2+5x+6) as x approaches -2. The attempt at solving the problem involves factoring and analyzing the limit of the numerator and denominator. It is determined that the limit does not exist. The conversation also briefly mentions using L'Hopital's rule, but it is not applicable in this scenario.
  • #1
xiaochobitz
6
0

Homework Statement





Homework Equations


[tex]
\lim(\frac{(x^2)-2x-3}{(x^2)+5x+6})[/tex]
[tex]
x->(-2)
[/tex]



The Attempt at a Solution




anyone can help out on this one?
 
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  • #2
Factor the quadratics. Which factor is going to zero? Can you cancel it?
 
  • #3
er by factoring it i actually get

[tex]

\lim(\frac{(x-3)(x+1)}{(x+3)(x+2)})[/tex]
[tex]

x->(-2)
[/tex]

am stuck from here on~
 
  • #4
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
tends to infinity?
 
  • #6
xiaochobitz said:
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'.
 
  • #7
so there is a problem in the question? haha~
 
  • #8
xiaochobitz said:
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.
 
  • #9
hmm icic~ troublesome question that got me stuck for so long haha~anyways, appreciate the help given, thanks so much~
 
  • #10
L'Hopitals?

( 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.
 
Last edited:
  • #11
whitay said:
L'Hopitals?

( 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.
 
  • #12
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.
 

FAQ: Limits for Quadratic Equations

What is the definition of a limit for a quadratic equation?

A limit for a quadratic equation is the value that a function approaches as its independent variable (usually denoted as x) gets closer and closer to a certain value. It represents the behavior of the function at that particular point.

How do you find the limit of a quadratic equation?

To find the limit of a quadratic equation, you can either use algebraic manipulation or graphing. Algebraically, you can substitute the value the function is approaching into the equation and solve for the limit. Graphically, you can plot the function and observe the behavior of the curve as the independent variable approaches the desired value.

What is the significance of finding the limit of a quadratic equation?

Finding the limit of a quadratic equation can help determine the behavior of the function at a certain point. It can also be used to find the maximum or minimum value of the function, as the limit at that point represents the highest or lowest point on the curve.

How does the degree of a quadratic equation affect its limit?

The degree of a quadratic equation does not affect its limit. As the independent variable approaches a certain value, the function will still approach a fixed value. However, the degree can affect the speed at which the function approaches the limit, which can be observed by graphing the function.

Are there any restrictions when finding the limit of a quadratic equation?

Yes, there are restrictions when finding the limit of a quadratic equation. The function must be continuous at the point where the limit is being evaluated. This means that the function must have a defined value at that point and the left and right limits must be equal. If this condition is not met, the limit does not exist.

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