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.
 A: Hint: 
$$\frac{\sin(x^2-1)}{x-1} = (x+1)\frac{\sin(x^2-1)}{x^2-1}$$
A: \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}
A: As $x$ approaches $1$, we have by Taylor's theorem that $\sin(x^2 - 1) = x^2 - 1 + O(x^2 -1)^3$, so we get that
$$\lim_{x\to 1} \frac{\sin(x^2-1)}{x-1} = \lim_{x\to 1} \frac{x^2-1+ O(x^2 -1)^3}{x-1}$$
$$= \lim_{x\to 1} x+1 +O(x^2 -1)^2(x+1) = 2$$
as the big-oh term goes to $0$.
A: It was just stop to think for a while that the answer came to me, dividing and multiplying the nominator by $x^2-1$ will give me $1$ from the fundamental trig limit times $x^2-1$.
So it will be $\frac{x^2-1}{x-1}$, factoring the nominator and dividing it by $x-1$ will have just $x+1$, which is equal to $2$.
A: $$
\lim_{x\to 1}\frac{\sin(x^2-1)}{x-1}=\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}=f'(1)
$$
where $f(x)=\sin(x^2-1)$. The derivative is found using the chain rule:
$$
f'(x)=\cos(x^2-1)\cdot 2x
$$
and hence $f'(1)=2$.
