# limit

• September 12th 2011, 12:20 PM
Olden
limit
trying to solve lim x->1 x^3 - 2x +1 / x-1
I see this gives 0/0. Now my problem starts, i don't know what to do when i have x^3..
• September 12th 2011, 12:21 PM
anonimnystefy
Re: limit
hi Olden

for this problem you will need l'Hospital's rule.do you know it?
• September 12th 2011, 12:26 PM
Olden
Re: limit
I found it in my book but weird since we are not at that chapter yet. I'm at chapter 2.2 and l'Hospital's rule is chapter 5.8... :/
• September 12th 2011, 12:29 PM
anonimnystefy
Re: limit
oh,i forgot.try dividing the numerator by the denominator.that should give an expression defined for x=1.
• September 12th 2011, 12:37 PM
Olden
Re: limit
I thought when i did not have lim x->infinity that i had to change the top x^3 - 2x +1 into something else. Like if i had lim x->2 x^2-2x / x^2-3x+2 = 0/0 i would make lim x->2 x(x-2) /(x-2)(x-1) = x/x-1 = 2. And i don't know how to do this when i have x^3. Sorry if i explain badly
• September 12th 2011, 12:40 PM
anonimnystefy
Re: limit
notice that x^3 - 2x +1 = (x-1)(x^2+x-1)
• September 12th 2011, 12:40 PM
Plato
Re: limit
Quote:

Originally Posted by Olden
I thought when i did not have lim x->infinity that i had to change the top x^3 - 2x +1 into something else. Like if i had lim x->2 x^2-2x / x^2-3x+2 = 0/0 i would make lim x->2 x(x-2) /(x-2)(x-1) = x/x-1 = 2. And i don't know how to do this when i have x^3. Sorry if i explain badly

$x^3-2x+1=(x-1)(x^2+x-1)$ now divide out $x-1$
• September 13th 2011, 12:13 PM
HallsofIvy
Re: limit
Note, by the way, that if P(a)= 0 for a polynomial P, then P must have a factor of (x-a). Any time you have a fraction of the form $\frac{P(x)}{Q(x)}$ and both P(a)= 0 and Q(a)= 0, then you can always factor out (x- a) from each and cancel it.