1. ## Finding Limits

Hi I started limits today and I'm having trouble with two problems. Any help would be appreciated.

The first problem is
lim = -2x - 4 / x^3 + 2x^2
x->-2

The 2nd problem is
lim = x^4 - 1 / x^3 - 1
x->1

Thanks guys. These two problems have been killing me for the last two hours. =(

2. Originally Posted by Afterme
The first problem is
lim = -2x - 4 / x^3 + 2x^2
x->-2
Factor $-2(x+2)/x^2(x+2)$ and cancel.

The 2nd problem is
lim = x^4 - 1 / x^3 - 1
x->1
Factor $(x-1)(x^3+x^2+x+1)/(x-1)(x^2+x+1)$ and cancel.

3. omg thanks. I can't believe I forgot something so simple . Thank you for helping me.

4. remember, your goal here is to simplify the function so that you can possibly plug in the number x is tending to. if this is not possible, then we have to deal with things differently. however, in both these cases, it is possible
Originally Posted by Afterme
Hi I started limits today and I'm having trouble with two problems. Any help would be appreciated.

The first problem is
lim = -2x - 4 / x^3 + 2x^2
x->-2
hint: factorzie

note that $\frac {-2x - 4}{x^3 + 2x^2} = \frac {-2(x + 2)}{x^2(x + 2)}$

The 2nd problem is
lim = x^4 - 1 / x^3 - 1
x->1

Thanks guys. These two problems have been killing me for the last two hours. =(
hint: the numerator is the difference of two squares and the denominator is the difference of two cubes

so, $\frac {x^4 - 1}{x^3 - 1} = \frac {(x^2 - 1)(x^2 + 1)}{(x - 1)(x^2 + x + 1)} = \frac {(x + 1)(x - 1)(x^2 + 1)}{(x - 1)(x^2 + x + 1)}$

EDIT: argh! and i am late again