Simplify Your Limit Problem with Basic Algebra Techniques"

[BoogieL80]

Okay, basic question. I'm working a limit problem and trying to factor part of my fraction:

t³ + 3t² - 12t + 4

Isn't there a way to combine like terms or something? I feel embarrassed asking this lol.

 

[fourier jr]

whats the original problem & how did you get that polynomial? maybe you don't really have to factor it...

 

[berkeman]

One (brute force) way is to write the factored form, multiply it out, and see if you can solve the simultaneous equations:

[tex]t^3 + 3t^2 - 12t + 4 = (t+a)(t+b)(t+c) + d[/tex]

 

[BoogieL80]

t³ + 3t² - 12t + 4 / t³ - 4t , as the limit of t approaches 2

I figured out that that I can get my denominator all the day down to t (t + 2)(t - 2), so I figured that there must be a way to pull that factor out of the numerator.

 

[0rthodontist]

So you're looking for a factor of either (t+2) or (t-2). Use polynomial division.

 

[berkeman]

Oh, it's a limit question. You call it a basic algebra question, but you may need to apply L'Hospital's (sp?) rule. That's beginning differential calculus. Is that what you're studying? I'll try to scare up a wikipedia entry on it...

 

[berkeman]

Yup, I spelled it wrong. Here's wikipedia's entry:

http://en.wikipedia.org/wiki/L'Hopital's_rule

Quiz question -- why would you normally want to apply L'Hopital's rule to this particular limit question?

 

[berkeman]

Hmm, that's interesting. Maybe you don't need it after all. That's neat how factoring the denomnator gets away from the infinity... I hadn't seen that before.

 

[berkeman]

Hmm, that's interesting. Maybe you don't need it after all. That's neat how factoring the denomnator gets away from the infinity... I hadn't seen that before.

In fact, once you factor the bottom to get rid of the infinity, can't you answer the limit question without factoring the numerator?

 

[BoogieL80]

That's a little ahead of where we are in my class, but thank you for the info ;)

 

[berkeman]

In fact, once you factor the bottom to get rid of the infinity, can't you answer the limit question without factoring the numerator?

Doh! I see why you need to factor the numerator still -- got to get rid of that t-2 term to avoid the infinity.

 

[BoogieL80]

I think so...I'm actually working on it now lol.

 

[matt grime]

In fact, once you factor the bottom to get rid of the infinity, can't you answer the limit question without factoring the numerator?

No. Factoring the bottom does not get rid of anything until you know it cancels with something in the numerator, so factor the numerator. Since we know one of the factors this is easy if we know polynomial division.

NB, I am assuming that both numerator and denominator vanish at the relevant point otherwise there is no need to do anything to find the limit.

 

[HallsofIvy]

As Orthodontist said before: the only reason you want to factor the numerator is to cancel the (x-2) in the denominator so assume that is a factor in the numerator. Just divide the numerator by x-2.