1. limit in calculus

I'm taking a calculus course in first year of university. There's a limit in my text I cannot seem to solve. Here it is, lim x to 0 of [(x+1)^(1/3) - 1]/x. The answers at the back say the limit is 1/3, but how would you re-arrange it to get this? NOTE: I cannot use L'Hopital's Rule. Thanks.

2. Originally Posted by BenWong
I'm taking a calculus course in first year of university. There's a limit in my text I cannot seem to solve. Here it is, lim x to 0 of [(x+1)^(1/3) - 1]/x. The answers at the back say the limit is 1/3, but how would you re-arrange it to get this? NOTE: I cannot use L'Hopital's Rule. Thanks.
Multiply this out.
$\frac{{\sqrt[3]{{x + 1}} - 1}}{x}\left( {\frac{{\sqrt[3]{{\left( {x + 1} \right)^2 }} + \sqrt[3]{{x + 1}} + 1}}{{\sqrt[3]{{\left( {x + 1} \right)^2 }} + \sqrt[3]{{x + 1}} + 1}}} \right)$

3. More easy:

Substitute $u^3=x+1.$

4. This one is similar. lim x to 0 of (sqrt(1+x) - sqrt(1-x))/x. I rationalized it to get 2/[sqrt(1+x) - sqrt(1-x)], but this doesn't make it any better. What do I do now?

5. Originally Posted by BenWong
This one is similar. lim x to 0 of (sqrt(1+x) - sqrt(1-x))/x. I rationalized it to get 2/[sqrt(1+x) $\color{red}+$ sqrt(1-x)], that should've been positive.

6. Yeah, that last one was a dumb mistake. Last one I don't know.
lim x to 1 of (x^2 - 3x + 2)/(x^3 - 3x^2 + 2).

7. All you need to do, it's to factorise top & bottom, 'cause there's a factor which produces the indetermination.