Thread: Prove that f has a limit at 1

1. Prove that f has a limit at 1

Ugh more review sheets!

Define: f0,1) -> R by f(x) = ($\displaystyle x^3$-$\displaystyle x^2$+x-1)/(x-1). Prove that f has a limit at 1.

It says that we shouldn't have to use $\displaystyle epsilon$ and $\displaystyle delta$ so that blew my whole idea of showing it was <$\displaystyle epsilon$. So would I got about it by dividing everything by $\displaystyle x^3$ and so on? or do I need to show an actual proof?

thanks

2. Originally Posted by tn11631
Ugh more review sheets!

Define: f0,1) -> R by f(x) = ($\displaystyle x^3$-$\displaystyle x^2$+x-1)/(x-1). Prove that f has a limit at 1.

It says that we shouldn't have to use $\displaystyle epsilon$ and $\displaystyle delta$ so that blew my whole idea of showing it was <$\displaystyle epsilon$. So would I got about it by dividing everything by $\displaystyle x^3$ and so on? or do I need to show an actual proof?

thanks

Factor the numerator

$\displaystyle x^3-x^2+x-1=x^2(x-1)+(x-1)=(x-1)(x^2+1)$

3. Originally Posted by TheEmptySet
Factor the numerator

$\displaystyle x^3-x^2+x-1=x^2(x-1)+(x-1)=(x-1)(x^2+1)$
so then I just plug in 1 from the f: (0,1) -> R , which would make the limit zero at 1? and I don't have to worry about 0 from f: (0,1), and I don't have to worry about a fancy proof? lol sorry I just want to make sure

4. Originally Posted by TheEmptySet
Factor the numerator

$\displaystyle x^3-x^2+x-1=x^2(x-1)+(x-1)=(x-1)(x^2+1)$
Duh I had this wrong..lol After we factor it we end up with ($\displaystyle x^2$+1) b.c every thing else cancels so then the limit at 1 is 2..lol thanks for the help!