Solve the limit as $x$ approaches $1$.
$$\lim_{x\rightarrow 1} \frac{\sin |x-1|}{x-1} $$
I know that $x$ approaches $1$ from negative side and positive side but I don't know where to start it. I tried to cancel out the $x-1$ on the bottom but did not work.
Thanks.

Hints: Using $\lim_{x \to 0}\frac{\sin x}{x}=1$, we have 
$$\lim_{ x\to 1^+}\frac{\sin|x-1|}{x-1}=\lim_{x\to 1^+}\frac{\sin(x-1)}{x-1}=1.$$
$$\lim_{ x\to 1^-}\frac{\sin|x-1|}{x-1}=\lim_{x\to 1^-}-\frac{\sin(x-1)}{x-1}=-1.$$
So its limit doesn't exist!
