You should be getting 3 as your answer (Sorry, 0 was a typo, I had 0 up here earlier, so your answer is correct).

If you actually take the derivative (which I know you can't use right now) you get:

$\displaystyle \frac{d}{dx}(x^3) = 3x^2\,dx = 3(1)^2 = 3$

Now, to do it without derivatives, you need to factor the top of this:

$\displaystyle \begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{x_1^3 - 1^3}{x_1 - 1}$

Now to factor two cubes (this is for any case):

$\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

So what you have is:

$\displaystyle \begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{x_1^3 - 1^3}{x_1 - 1} =

\begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{(x_1 - 1)(x_1^2 + x_1 + 1)}{x_1 - 1} =

\begin{array}{c}lim\\x_1\rightarrow 1\end{array} (x_1^2 + x_1 + 1) = 3

$

Sorry, I had a few typos

. But your answer of 3 is correct. Hopefully all my work is correct now. Sorry again for any confusion.