Limit of trig function How to simplify this limit? 1 hour already and I haven't figured out a way yet.
$$\lim_{x\to 1} \frac{\sin(x^2-1)}{x-1}$$
Withou using l'hôpital's rule, please.

\begin{align}
\lim_{x\to 1} \dfrac{\sin(x^2-1)}{x-1}=&\lim_{x\to 1} \dfrac{\sin(x^2-1)}{x-1}\times\dfrac{x+1}{x+1}\\=&\lim_{x^2-1\to 0} \dfrac{\sin(x^2-1)}{x^2-1}\times\lim_{x\to1}(x+1)\\=&1\times(1+1)\\=&2
\end{align}
