Find the limit without l'Hôpital's rule Find the limit
$$\lim_{x\to 1}\frac{(x^2-1)\sin(3x-3)}{\cos(x^3-1)\tan^2(x^2-x)}.$$
I'm a little rusty with limits, can somebody please give me some pointers on how to solve this one? Also, l'Hôpital's rule isn't allowed in case you were thinking of using it. Thanks in advance.

The solution below is the same as the solution by Marvis. Note that as $x\to 1$, $\cos(x^2-1)\to 1$, so we can forget about it. Also, $x^2-1=(x+1)(x-1)$, and the $x+1$ part approaches $2$, so we can sort of forget about it, multiplying by $2$ at the end.  And $\tan^2(x^2-x)=\dfrac{\sin^2 (x^2-x)}{\cos^2(x^2-x)}$. The $\cos^2(x^2-x)$ part harmlessly approaches $1$.
So we are interested only in 
$$\lim_{x\to 1}\frac{(x-1)\sin(3x-3)}{\sin^2(x^2-x)}.$$
I feel better already. Don't really like all those $x-1$, so let $u=x-1$. Then
$x^2-x=u(u+1)$ and we are down to 
$$\lim_{u\to 0}\frac{u\sin 3u}{\sin^2(u(u+1))}.\tag{$1$}$$
Multiply top and bottom by $3(u(u+1))^2$. After minor cancellaton, we want 
$$\lim_{u\to 0} \frac{3}{(u+1)^2}\cdot\frac{\sin 3u}{3u}\cdot \frac{(u(u+1))^2}{\sin^2(u(u+1))}.$$
We are down to three terms, the first of which has limit $3$, and the last two of which are variants of the standard $\lim_{t\to 0}\dfrac{\sin t}{t}$. So the limit is $3$. Finally, we remember about the factor $2$ that we jettisoned earlier. 
Remark: After we reach Expression $1$, the multiplying top and bottom by $3(u(u+1))^2$ is just to please the grader. Near $0$, $\sin t$ and $t$ are first-order indistinguishable, so $\sin 3u\approx 3u$ and $\sin^2 (u(u+1))\approx u^2(u+1)^2$. Indeed, $\tan t$ and $t$ are also first-order indistinguishable, and one can read off the limit without any calculation. 
