# Another limit with l'hospital

1. May 19, 2010

### Asphyxiated

1. The problem statement, all variables and given/known data

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

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 whats 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?

2. May 19, 2010

### 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)

3. May 19, 2010

### 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.

4. May 19, 2010

### gabbagabbahey

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