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.

$$\dfrac{\sin(3x-3)}{\tan^2(x^2-x)} = \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x-1)}{x^2(x-1)^2}$$
Hence, $$\dfrac{(x^2-1)\sin(3x-3)}{\cos(x^3-1)\tan^2(x^2-x)} = \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x-1)(x^2-1)}{x^2(x-1)^2 \cos(x^3-1)}\\
 = \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x+1)}{x^2 \cos(x^3-1)}$$
Now the first and second term on the right has limit $1$ as $x \to 1$. The last term limit can be obtained by plugging $x=1$, to give the limit as $$1 \times 1 \times \dfrac{3 \times (1+1)}{1^2 \times \cos(0)} = 6$$
