What is the next step for a 0/0 limit with l'Hospital's rule?

[Asphyxiated]

Homework Statement

$$\lim_{u \to 1} \frac {(u-1)^{3}}{u^{-1}-u^{2}+3u-3}$$

The Attempt at a Solution

So the first derivative is:

$$\lim_{u \to 1} \frac {3(u-1)^{2}}{-u^{-2}-2u+3}$$

this limit also gives a 0/0 situation so take the derivative again:

$$\lim_{u \to 1} \frac{6(u-1)}{2u^{-3}-2}$$

and this limit as well gives a 0/0 again, so would I do it again? or what's the next step? The book doesn't give me any instruction to this situation just to the 2nd derivative.

If I do take the derivative again it is -1, is that correct?

 

[gabbagabbahey]

There's no need to use L'Hopital's rule here. Just start by multiplying both numerator and denominator by $u$, and then factor the denominator (you already know $u=1$ is a factor, so it should be easy)

 

[Asphyxiated]

right well its from the section in my book to learn l'hopital's rule, so that is why I am using it here.

 

[gabbagabbahey]

In that case, yes, just differentiate one more time to get -1.