Find $\lim_{x\rightarrow 1}\frac{\sin{(x^2-1)}}{x-1}$ My problem is to calculate
$$\lim_{x\rightarrow 1}\frac{\sin{(x^2-1)}}{x-1}$$
I evaluated to $\frac{\sin(x+1)(x-1)}{x-1}$ and then to $\sin(x+1)$ but I wonder about the result of limit as x approach 1 of $\sin{(x+1)}$. I cannot think of any result I cannot connect to anything.

Notice
$$ \frac{\sin ( x^2 - 1 )}{x-1} = \frac{\sin ( x^2 - 1 )}{x-1} \cdot \frac{ x+1}{x+1} = \frac{\sin ( x^2 - 1 )}{x^2-1} \cdot (x+1) \to_{x \to 1} 2 $$
since 
$$ \frac{ \sin y}{y} \to_{y \to 0} 1 $$
